Integration of rational function by using partial fractions, Questions in English

(C) To evaluate the integral $\int \frac{dx}{1 - x^2}$,we use the method of partial fractions.
The integrand can be written as $\frac{1}{1 - x^2} = \frac{1}{(1 - x)(1 + x)}$.
Using partial fractions,we have $\frac{1}{(1 - x)(1 + x)} = \frac{A}{1 - x} + \frac{B}{1 + x}$.
Solving for $A$ and $B$,we get $1 = A(1 + x) + B(1 - x)$.
Setting $x = 1$,we get $1 = 2A \implies A = \frac{1}{2}$.
Setting $x = -1$,we get $1 = 2B \implies B = \frac{1}{2}$.
Thus,$\int \frac{dx}{1 - x^2} = \int \left( \frac{1/2}{1 - x} + \frac{1/2}{1 + x} \right) dx$.
Integrating term by term,we get $\frac{1}{2} \int \frac{dx}{1 - x} + \frac{1}{2} \int \frac{dx}{1 + x} = -\frac{1}{2} \ln |1 - x| + \frac{1}{2} \ln |1 + x| + c$.
Using the property of logarithms,$\ln a - \ln b = \ln(a/b)$,we get $\frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| + c$.

(A) To evaluate the integral $I = \int \frac{dx}{x - x^2}$,we first factor the denominator:
$x - x^2 = x(1 - x)$.
Using partial fraction decomposition,we write:
$\frac{1}{x(1 - x)} = \frac{A}{x} + \frac{B}{1 - x}$.
Multiplying by $x(1 - x)$,we get $1 = A(1 - x) + Bx$.
Setting $x = 0$,we find $A = 1$.
Setting $x = 1$,we find $B = 1$.
Thus,$\frac{1}{x(1 - x)} = \frac{1}{x} + \frac{1}{1 - x}$.
Integrating both sides with respect to $x$:
$I = \int \left( \frac{1}{x} + \frac{1}{1 - x} \right) dx$.
$I = \log |x| - \log |1 - x| + C$ (since the derivative of $1 - x$ is $-1$).
Therefore,the correct option is $A$.

(A) We have the integral $I = \int \frac{dx}{1 + x + x^2 + x^3}$.
Factor the denominator: $1 + x + x^2 + x^3 = (1 + x) + x^2(1 + x) = (1 + x)(1 + x^2)$.
Using partial fractions: $\frac{1}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$.
$1 = A(1 + x^2) + (Bx + C)(1 + x)$.
For $x = -1$,$1 = A(1 + 1) \implies A = \frac{1}{2}$.
Comparing coefficients: $A + B = 0 \implies B = -\frac{1}{2}$,and $A + C = 1 \implies C = \frac{1}{2}$.
So,$I = \int (\frac{1/2}{1 + x} + \frac{-1/2 x + 1/2}{1 + x^2}) dx = \frac{1}{2} \int \frac{dx}{1 + x} - \frac{1}{4} \int \frac{2x}{1 + x^2} dx + \frac{1}{2} \int \frac{dx}{1 + x^2}$.
$I = \frac{1}{2} \log|1 + x| - \frac{1}{4} \log(1 + x^2) + \frac{1}{2} \tan^{-1} x + c$.
$I = \log \sqrt{1 + x} - \frac{1}{2} \log \sqrt{1 + x^2} + \frac{1}{2} \tan^{-1} x + c$.

(B) To solve the integral $\int \frac{x - 1}{(x - 3)(x - 2)} \, dx$,we use partial fractions. $\frac{x - 1}{(x - 3)(x - 2)} = \frac{A}{x - 3} + \frac{B}{x - 2}$.
Multiplying by $(x - 3)(x - 2)$,we get $x - 1 = A(x - 2) + B(x - 3)$.
For $x = 3$,$3 - 1 = A(3 - 2) \Rightarrow A = 2$.
For $x = 2$,$2 - 1 = B(2 - 3) \Rightarrow B = -1$.
Thus,$\int \left( \frac{2}{x - 3} - \frac{1}{x - 2} \right) \, dx = 2 \log |x - 3| - \log |x - 2| + c$.
Using logarithmic properties,this is $\log |(x - 3)^2| - \log |x - 2| + c$.

(C) To evaluate the integral $\int \frac{dx}{(x + 1)(x + 2)}$,we use partial fractions.
We can write the integrand as:
$\frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$
Multiplying both sides by $(x + 1)(x + 2)$,we get:
$1 = A(x + 2) + B(x + 1)$
Setting $x = -1$,we get $A = 1$.
Setting $x = -2$,we get $B = -1$.
Thus,$\int \frac{dx}{(x + 1)(x + 2)} = \int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) dx$
$= \log |x + 1| - \log |x + 2| + C$
$= \log \left| \frac{x + 1}{x + 2} \right| + C$.

(A) To evaluate the integral $\int \frac{x}{(x - 2)(x - 1)} \, dx$,we use partial fraction decomposition.
Let $\frac{x}{(x - 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}$.
Multiplying by $(x - 2)(x - 1)$,we get $x = A(x - 1) + B(x - 2)$.
Setting $x = 1$,we get $1 = B(1 - 2) \implies B = -1$.
Setting $x = 2$,we get $2 = A(2 - 1) \implies A = 2$.
Thus,$\int \frac{x}{(x - 2)(x - 1)} \, dx = \int \left( \frac{2}{x - 2} - \frac{1}{x - 1} \right) \, dx$.
Integrating term by term,we get $2 \log_e |x - 2| - \log_e |x - 1| + p$.
Using logarithmic properties,this simplifies to $\log_e \frac{(x - 2)^2}{|x - 1|} + p$.

(A) We use partial fractions: $\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.
$1 = A(x^2 + 1) + (Bx + C)(x - 1)$.
Setting $x = 1$,we get $1 = A(1 + 1) \Rightarrow A = \frac{1}{2}$.
Comparing coefficients of $x^2$: $A + B = 0 \Rightarrow B = -\frac{1}{2}$.
Comparing constant terms: $A - C = 1 \Rightarrow C = A - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$.
Thus,$\frac{1}{(x - 1)(x^2 + 1)} = \frac{1}{2(x - 1)} - \frac{x + 1}{2(x^2 + 1)}$.
Integrating both sides: $\int \frac{1}{2(x - 1)} dx - \frac{1}{2} \int \frac{x}{x^2 + 1} dx - \frac{1}{2} \int \frac{1}{x^2 + 1} dx$.
$= \frac{1}{2}\log |x - 1| - \frac{1}{4}\log (x^2 + 1) - \frac{1}{2}\tan^{-1} x + c$.

(B) We have the integral $I = \int {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} \,dx$.
First,perform polynomial division: $\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}} = \frac{({x^2} + x - 6) + 5}{{{x^2} + x - 6}} = 1 + \frac{5}{{{x^2} + x - 6}}$.
Now,factor the denominator: ${x^2} + x - 6 = (x + 3)(x - 2)$.
Using partial fractions: $\frac{5}{{(x + 3)(x - 2)}} = \frac{A}{x + 3} + \frac{B}{x - 2}$.
$5 = A(x - 2) + B(x + 3)$.
Setting $x = 2$,we get $5 = 5B \implies B = 1$.
Setting $x = -3$,we get $5 = -5A \implies A = -1$.
Thus,$\frac{5}{{(x + 3)(x - 2)}} = \frac{1}{x - 2} - \frac{1}{x + 3}$.
Integrating: $I = \int {\left( {1 + \frac{1}{x - 2} - \frac{1}{x + 3}} \right)} \,dx = x + \log |x - 2| - \log |x + 3| + c$.

(B) Let $I = \int {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} \,dx$.
Substitute $t = {x^2}$,then the partial fraction decomposition of $\frac{t}{(t+2)(t+3)}$ is $\frac{A}{t+2} + \frac{B}{t+3}$.
$t = A(t+3) + B(t+2)$.
For $t = -2$,$-2 = A(1) \implies A = -2$.
For $t = -3$,$-3 = B(-1) \implies B = 3$.
So,$\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}} = \frac{3}{{{x^2} + 3}} - \frac{2}{{{x^2} + 2}}$.
Integrating both sides,$I = \int {\frac{3}{{{x^2} + 3}}} \,dx - \int {\frac{2}{{{x^2} + 2}}} \,dx$.
Using the formula $\int {\frac{1}{{{x^2} + {a^2}}}} \,dx = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + c$,we get:
$I = 3 \cdot \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} - 2 \cdot \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + c$.
$I = \sqrt 3 {\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} - \sqrt 2 {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + c$.

(C) Let $I = \int \frac{dx}{(x^2 + 1)(x^2 + 4)}$.
Using partial fractions,let $t = x^2$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.
$1 = A(t+4) + B(t+1)$.
For $t = -1$,$1 = 3A \Rightarrow A = \frac{1}{3}$.
For $t = -4$,$1 = -3B \Rightarrow B = -\frac{1}{3}$.
So,$\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3} \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} \right)$.
Integrating both sides:
$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.
Using the formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{3} \tan^{-1}x - \frac{1}{3} \left( \frac{1}{2} \tan^{-1}\frac{x}{2} \right) + c$.
$I = \frac{1}{3} \tan^{-1}x - \frac{1}{6} \tan^{-1}\frac{x}{2} + c$.

(D) To evaluate the integral $I = \int \frac{1}{x - x^3} \, dx$,we first factor the denominator:
$x - x^3 = x(1 - x^2) = x(1 - x)(1 + x)$.
Using partial fractions:
$\frac{1}{x(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x}$.
Solving for constants,we get $A = 1$,$B = 1/2$,and $C = -1/2$.
Thus,$\frac{1}{x - x^3} = \frac{1}{x} + \frac{1}{2(1 - x)} - \frac{1}{2(1 + x)} = \frac{1}{x} + \frac{1}{2} \left( \frac{1}{1 - x} - \frac{1}{1 + x} \right)$.
Integrating term by term:
$I = \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{1 - x} \, dx - \frac{1}{2} \int \frac{1}{1 + x} \, dx$.
$I = \log |x| - \frac{1}{2} \log |1 - x| - \frac{1}{2} \log |1 + x| + c$.
$I = \frac{1}{2} [2 \log |x| - \log |1 - x| - \log |1 + x|] + c$.
$I = \frac{1}{2} \log \left| \frac{x^2}{(1 - x)(1 + x)} \right| + c = \frac{1}{2} \log \frac{x^2}{1 - x^2} + c$.

(C) Let $I = \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx$.
Substitute $\sin x = t$,then $\cos x \,dx = dt$.
The integral becomes $I = \int \frac{dt}{(t + 1)(t + 2)}$.
Using partial fractions,$\frac{1}{(t + 1)(t + 2)} = \frac{1}{t + 1} - \frac{1}{t + 2}$.
Integrating both terms,$I = \int \frac{1}{t + 1} \,dt - \int \frac{1}{t + 2} \,dt = \log |t + 1| - \log |t + 2| + c$.
Using the property $\log a - \log b = \log \frac{a}{b}$,we get $I = \log \left| \frac{t + 1}{t + 2} \right| + c$.
Substituting $t = \sin x$ back,$I = \log \left| \frac{1 + \sin x}{2 + \sin x} \right| + c$.

(B) Let $I = \int \frac{e^x}{(1 + e^x)(2 + e^x)} dx$.
Using partial fractions,let $\frac{e^x}{(1 + e^x)(2 + e^x)} = \frac{A}{1 + e^x} + \frac{B}{2 + e^x}$.
Then $e^x = A(2 + e^x) + B(1 + e^x)$.
Setting $e^x = -1$,we get $A = -1$.
Setting $e^x = -2$,we get $B = 1$.
So,the integral becomes $\int \left( \frac{1}{2 + e^x} - \frac{1}{1 + e^x} \right) e^x dx$.
Let $u = e^x$,then $du = e^x dx$.
The integral is $\int \left( \frac{1}{2 + u} - \frac{1}{1 + u} \right) du = \log|2 + u| - \log|1 + u| + c = \log \left| \frac{2 + e^x}{1 + e^x} \right| + c$.
Note: The provided option $B$ is $\log \left( \frac{1 + e^x}{2 + e^x} \right) + c$,which is $-\log \left| \frac{2 + e^x}{1 + e^x} \right| + c$. Since the constant $c$ is arbitrary,the result is equivalent to $\log \left| \frac{1 + e^x}{2 + e^x} \right| + c$.

(C) Let $I = \int \frac{dx}{e^x + 1 - 2e^{-x}}$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x dx}{e^{2x} + e^x - 2}$.
Let $e^x = t$,then $e^x dx = dt$.
$I = \int \frac{dt}{t^2 + t - 2} = \int \frac{dt}{(t + 2)(t - 1)}$.
Using partial fractions:
$\frac{1}{(t + 2)(t - 1)} = \frac{1}{3} \left[ \frac{1}{t - 1} - \frac{1}{t + 2} \right]$.
Integrating both sides:
$I = \frac{1}{3} \int \left( \frac{1}{t - 1} - \frac{1}{t + 2} \right) dt = \frac{1}{3} \log |t - 1| - \frac{1}{3} \log |t + 2| + c$.
Substituting $t = e^x$ back:
$I = \frac{1}{3} \log (e^x - 1) - \frac{1}{3} \log (e^x + 2) + c$.

(C) Let $I = \int \frac{x \, dx}{(x^2 - a^2)(x^2 - b^2)}$.
Put $x^2 = t$,then $2x \, dx = dt$,or $x \, dx = \frac{dt}{2}$.
Substituting this into the integral,we get:
$I = \int \frac{dt}{2(t - a^2)(t - b^2)} = \frac{1}{2} \int \frac{dt}{(t - a^2)(t - b^2)}$.
Using partial fractions:
$\frac{1}{(t - a^2)(t - b^2)} = \frac{1}{a^2 - b^2} \left( \frac{1}{t - a^2} - \frac{1}{t - b^2} \right)$.
Therefore,$I = \frac{1}{2(a^2 - b^2)} \int \left( \frac{1}{t - a^2} - \frac{1}{t - b^2} \right) dt$.
$I = \frac{1}{2(a^2 - b^2)} [\log |t - a^2| - \log |t - b^2|] + c$.
$I = \frac{1}{2(a^2 - b^2)} \log \left| \frac{t - a^2}{t - b^2} \right| + c$.
Substituting $t = x^2$ back,we get:
$I = \frac{1}{2(a^2 - b^2)} \log \left| \frac{x^2 - a^2}{x^2 - b^2} \right| + c$.

(A) To evaluate the integral $I = \int \frac{1}{(x^2 + a^2)(x^2 + b^2)} dx$,we use partial fractions.
We can write the integrand as:
$\frac{1}{(x^2 + a^2)(x^2 + b^2)} = \frac{1}{a^2 - b^2} \left( \frac{1}{x^2 + b^2} - \frac{1}{x^2 + a^2} \right)$.
Now,integrate both sides with respect to $x$:
$I = \frac{1}{a^2 - b^2} \int \left( \frac{1}{x^2 + b^2} - \frac{1}{x^2 + a^2} \right) dx$.
Using the standard integral formula $\int \frac{1}{x^2 + k^2} dx = \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right) + c$,we get:
$I = \frac{1}{a^2 - b^2} \left[ \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) - \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \right] + c$.

(C) Let $I = \int \frac{dx}{x(x^n + 1)}$.
Multiply the numerator and denominator by $x^{n-1}$:
$I = \int \frac{x^{n-1} dx}{x^n(x^n + 1)}$.
Let $x^n = t$,then $n x^{n-1} dx = dt$,which implies $x^{n-1} dx = \frac{dt}{n}$.
Substituting these into the integral:
$I = \frac{1}{n} \int \frac{dt}{t(t+1)}$.
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
$I = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$.
$I = \frac{1}{n} (\log |t| - \log |t+1|) + c$.
$I = \frac{1}{n} \log \left| \frac{t}{t+1} \right| + c$.
Substituting $t = x^n$ back:
$I = \frac{1}{n} \log \frac{x^n}{x^n + 1} + c$.

(C) Let $I = \int \frac{dx}{x(x^5 + 1)}$.
Multiply the numerator and denominator by $x^4$:
$I = \int \frac{x^4 dx}{x^5(x^5 + 1)}$.
Let $x^5 = t$,then $5x^4 dx = dt$,which implies $x^4 dx = \frac{dt}{5}$.
Substituting these into the integral:
$I = \frac{1}{5} \int \frac{dt}{t(t + 1)}$.
Using partial fractions:
$\frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1}$.
Therefore,$I = \frac{1}{5} \int (\frac{1}{t} - \frac{1}{t + 1}) dt$.
$I = \frac{1}{5} (\log |t| - \log |t + 1|) + c$.
$I = \frac{1}{5} \log |\frac{t}{t + 1}| + c$.
Substituting $t = x^5$ back:
$I = \frac{1}{5} \log |\frac{x^5}{x^5 + 1}| + c$.
Comparing this with the given options,option $C$ is correct.

(A) Let $I = \int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}}} dx$.
Using partial fractions,let $\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}} = \frac{A}{{{x^2} - 1}} + \frac{B}{{{x^2} - 4}}$.
Then $2{x^2} + 3 = A({x^2} - 4) + B({x^2} - 1)$.
Comparing coefficients of ${x^2}$ and constant terms:
$A + B = 2$ $(i)$
$-4A - B = 3$ $(ii)$
Adding $(i)$ and $(ii)$,$-3A = 5 \implies A = -\frac{5}{3}$.
Substituting $A$ in $(i)$,$B = 2 + \frac{5}{3} = \frac{11}{3}$.
So,$I = \int {\frac{{ - 5/3}}{{{x^2} - 1}} dx} + \int {\frac{{11/3}}{{{x^2} - 4}} dx}$.
Using $\int {\frac{{dx}}{{{x^2} - {a^2}}}} = \frac{1}{{2a}}\log \left| {\frac{{x - a}}{{x + a}}} \right| + c$:
$I = -\frac{5}{3} \cdot \frac{1}{2} \log \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{{11}}{3} \cdot \frac{1}{4} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$.
$I = -\frac{5}{6} \log \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{{11}}{{12}} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$.
$I = \frac{5}{6} \log \left| {\frac{{x + 1}}{{x - 1}}} \right| + \frac{{11}}{{12}} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$.
Comparing with $\log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^a}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^b} + c$,we get $a = \frac{{11}}{{12}}$ and $b = \frac{5}{6}$.

(C) We are given the integral: $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx$.
First,we express the numerator in terms of the derivative of the denominator: $\frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$.
So,$\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5 + 8}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5}}{{{x^2} - 5x + 6}}} \;dx + \int {\frac{8}{{(x - 2)(x - 3)}}} \;dx$.
Using partial fractions for the second part: $\frac{8}{(x - 2)(x - 3)} = \frac{8}{x - 3} - \frac{8}{x - 2}$.
Thus,the integral becomes: $\ln |x^2 - 5x + 6| + 8 \int (\frac{1}{x - 3} - \frac{1}{x - 2}) \;dx + C$.
$= \ln |(x - 2)(x - 3)| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$.
$= \ln |x - 2| + \ln |x - 3| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$.
$= 9 \ln |x - 3| - 7 \ln |x - 2| + C$.
Comparing this with the given expression $9 \ln (x - 3) - 7 \ln (x - 2) + A$,we find that $A$ must be an arbitrary constant $C$.

(A) Let $I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{\sin x(1 + 2\cos x)}$.
Multiply numerator and denominator by $\sin x$:
$I = \int \frac{\sin x \, dx}{\sin^2 x(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos x)(1 + \cos x)(1 + 2\cos x)}$.
Let $\cos x = t$,then $-\sin x \, dx = dt$,so $\sin x \, dx = -dt$.
$I = - \int \frac{dt}{(1 - t)(1 + t)(1 + 2t)}$.
Using partial fractions:
$\frac{1}{(1 - t)(1 + t)(1 + 2t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{1 + 2t}$.
Solving for coefficients,we get $A = \frac{1}{6}$,$B = -\frac{1}{2}$,$C = \frac{4}{3}$.
$I = - \int \left( \frac{1/6}{1 - t} - \frac{1/2}{1 + t} + \frac{4/3}{1 + 2t} \right) dt$.
$I = - \left[ \frac{1}{6} \frac{\log |1 - t|}{-1} - \frac{1}{2} \log |1 + t| + \frac{4}{3} \cdot \frac{1}{2} \log |1 + 2t| \right] + C$.
$I = \frac{1}{6} \log |1 - t| + \frac{1}{2} \log |1 + t| - \frac{2}{3} \log |1 + 2t| + C$.
Substituting $t = \cos x$ back:
$I = \frac{1}{6} \log |1 - \cos x| + \frac{1}{2} \log |1 + \cos x| - \frac{2}{3} \log |1 + 2\cos x| + C$.

(D) Let $I = \int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx}$.
Using partial fractions,we write $\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}} = \frac{A_1}{x - 1} + \frac{{Bx + C}}{{{x^2} + 1}}$.
Solving for constants,we get $2x + 3 = A_1({x^2} + 1) + (Bx + C)(x - 1)$.
For $x = 1$,$5 = 2A_1 \implies A_1 = \frac{5}{2}$.
Comparing coefficients of $x^2$,$0 = A_1 + B \implies B = -\frac{5}{2}$.
Comparing constants,$3 = A_1 - C \implies C = A_1 - 3 = \frac{5}{2} - 3 = -\frac{1}{2}$.
Thus,$I = \int {\frac{{5/2}}{{x - 1}}dx} + \int {\frac{{-5/2x - 1/2}}{{{x^2} + 1}}dx}$.
$I = \frac{5}{2}\log |x - 1| - \frac{5}{4}\int {\frac{{2x}}{{1 + {x^2}}}dx} - \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}dx}$.
$I = \frac{5}{2}\log |x - 1| - \frac{5}{4}\log (1 + {x^2}) - \frac{1}{2}{\tan ^{ - 1}}x + A$.
$I = \log \left\{ {{{(x - 1)}^{\frac{5}{2}}}{{({x^2} + 1)}^{ - 5/4}}} \right\} - \frac{1}{2}{\tan ^{ - 1}}x + A$.
Comparing with the given expression,$a = -\frac{5}{4}$.

(A) Let $I = \int {\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}}dx}$.
Using partial fractions,we write $\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}} = \frac{A}{{x^2 - 4}} + \frac{B}{{x^2 - 1}}$.
$2{x^2} + 1 = A(x^2 - 1) + B(x^2 - 4)$.
For $x^2 = 1$,$3 = B(1 - 4) \implies 3 = -3B \implies B = -1$.
For $x^2 = 4$,$9 = A(4 - 1) \implies 9 = 3A \implies A = 3$.
So,$\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}} = \frac{3}{{x^2 - 4}} - \frac{1}{{x^2 - 1}}$.
$I = \int {\frac{3}{{x^2 - 2^2}}dx} - \int {\frac{1}{{x^2 - 1^2}}dx}$.
Using the formula $\int {\frac{1}{{x^2 - a^2}}dx = \frac{1}{{2a}}\log \left| {\frac{{x - a}}{{x + a}}} \right| + C}$,we get:
$I = 3 \cdot \frac{1}{{2(2)}}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| - \frac{1}{{2(1)}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| + C$.
$I = \frac{3}{4}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{1}{2}\log \left| {\frac{{x + 1}}{{x - 1}}} \right| + C$.
Comparing this with $\log \left[ {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^a}{{\left( {\frac{{x - 2}}{{x + 2}}} \right)}^b}} \right] + C$,we get $a = 1/2$ and $b = 3/4$.

(D) Let $I = \int_1^2 \frac{dx}{x(1 + x^4)}$.
Multiply the numerator and denominator by $x^3$:
$I = \int_1^2 \frac{x^3 dx}{x^4(1 + x^4)}$.
Let $x^4 = t$,then $4x^3 dx = dt$,which implies $x^3 dx = \frac{dt}{4}$.
When $x = 1$,$t = 1^4 = 1$. When $x = 2$,$t = 2^4 = 16$.
Substituting these into the integral:
$I = \frac{1}{4} \int_1^{16} \frac{dt}{t(1 + t)}$.
Using partial fractions,$\frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t}$.
$I = \frac{1}{4} \int_1^{16} \left( \frac{1}{t} - \frac{1}{1 + t} \right) dt$.
$I = \frac{1}{4} [\log|t| - \log|1 + t|]_1^{16} = \frac{1}{4} [\log|\frac{t}{1 + t}|]_1^{16}$.
$I = \frac{1}{4} (\log \frac{16}{17} - \log \frac{1}{2}) = \frac{1}{4} \log (\frac{16}{17} \times 2) = \frac{1}{4} \log \frac{32}{17}$.

(C) Let $I = \int \frac{dx}{1 - \cos x - \sin x}$.
Using half-angle formulas $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$ and $\sin x = \frac{2\tan(x/2)}{1 + \tan^2(x/2)}$,we get:
$I = \int \frac{dx}{1 - \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} - \frac{2\tan(x/2)}{1 + \tan^2(x/2)}}$
$I = \int \frac{(1 + \tan^2(x/2)) dx}{1 + \tan^2(x/2) - 1 + \tan^2(x/2) - 2\tan(x/2)}$
$I = \int \frac{\sec^2(x/2) dx}{2\tan^2(x/2) - 2\tan(x/2)}$
Let $t = \tan(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.
$I = \int \frac{2 dt}{2(t^2 - t)} = \int \frac{dt}{t(t - 1)}$
Using partial fractions: $\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}$.
$I = \int \left( \frac{1}{t - 1} - \frac{1}{t} \right) dt = \log|t - 1| - \log|t| + c = \log\left| \frac{t - 1}{t} \right| + c$
$I = \log\left| \frac{\tan(x/2) - 1}{\tan(x/2)} \right| + c = \log|1 - \cot(x/2)| + c$.

(A) We have $I = \int \frac{x^4}{(x-1)(x^2+1)} dx$.
Using polynomial division or algebraic manipulation:
$x^4 = (x^4-1) + 1 = (x^2-1)(x^2+1) + 1 = (x-1)(x+1)(x^2+1) + 1$.
So,$\frac{x^4}{(x-1)(x^2+1)} = (x+1) + \frac{1}{(x-1)(x^2+1)}$.
Now,we use partial fractions for $\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$.
$1 = A(x^2+1) + (Bx+C)(x-1)$.
Setting $x=1$,we get $1 = 2A \implies A = 1/2$.
Comparing coefficients: $x^2: A+B=0 \implies B = -1/2$.
Constant: $A-C=1 \implies C = A-1 = 1/2 - 1 = -1/2$.
Thus,$\frac{1}{(x-1)(x^2+1)} = \frac{1}{2(x-1)} - \frac{x+1}{2(x^2+1)} = \frac{1}{2(x-1)} - \frac{x}{2(x^2+1)} - \frac{1}{2(x^2+1)}$.
Integrating: $\int (x+1) dx + \frac{1}{2} \int \frac{1}{x-1} dx - \frac{1}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
$= \frac{x^2}{2} + x + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} \tan^{-1} x + c$.
$= \frac{x^2+2x}{2} + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} \tan^{-1} x + c = \frac{x(x+2)}{2} + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} \tan^{-1} x + c$.

(B) We have the integral $I = \int {\frac{{{x^3} - 1}}{{{x^3} + x}}dx}$.
First,simplify the integrand: $\frac{{{x^3} - 1}}{{{x^3} + x}} = \frac{{{x^3} + x - x - 1}}{{x({x^2} + 1)}} = \frac{{{x^3} + x}}{{x({x^2} + 1)}} - \frac{{x + 1}}{{x({x^2} + 1)}} = 1 - \frac{{x + 1}}{{x({x^2} + 1)}}$.
Using partial fractions for $\frac{{x + 1}}{{x({x^2} + 1)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}}$.
$x + 1 = A({x^2} + 1) + (Bx + C)x = (A + B){x^2} + Cx + A$.
Comparing coefficients,we get $A = 1$,$C = 1$,and $A + B = 0 \implies B = -1$.
So,$\frac{{x + 1}}{{x({x^2} + 1)}} = \frac{1}{x} - \frac{{x - 1}}{{{x^2} + 1}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}$.
Now,$I = \int {\left( {1 - \left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}}\right)} \right)dx} = \int {\left( {1 - \frac{1}{x} + \frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}}} \right)dx}$.
Integrating term by term: $I = x - \log |x| + \frac{1}{2}\log ({x^2} + 1) - {\tan ^{ - 1}}x + c$.
Since $\frac{1}{2}\log ({x^2} + 1) = \log \sqrt {{x^2} + 1}$,the result is $x - \log |x| + \log \sqrt {{x^2} + 1} - {\tan ^{ - 1}}x + c$.

(B) To solve the integral $I = \int \frac{1}{x(x^4 - 1)} \, dx$,we multiply the numerator and denominator by $x^3$:
$I = \int \frac{x^3}{x^4(x^4 - 1)} \, dx$.
Let $u = x^4$,then $du = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{du}{4}$.
Substituting these into the integral:
$I = \frac{1}{4} \int \frac{1}{u(u - 1)} \, du$.
Using partial fractions: $\frac{1}{u(u - 1)} = \frac{1}{u - 1} - \frac{1}{u}$.
$I = \frac{1}{4} \int \left( \frac{1}{u - 1} - \frac{1}{u} \right) \, du = \frac{1}{4} [\log |u - 1| - \log |u|] + K$.
$I = \frac{1}{4} \log \left| \frac{u - 1}{u} \right| + K$.
Substituting $u = x^4$ back:
$I = \frac{1}{4} \log \left( \frac{x^4 - 1}{x^4} \right) + K$.

(D) Let $I = \int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}}dx}$.
Using partial fractions,let $t = x^2$. Then $\frac{2t+3}{(t-1)(t+4)} = \frac{A}{t-1} + \frac{B}{t+4}$.
$2t+3 = A(t+4) + B(t-1)$.
For $t=1$,$5 = 5A \Rightarrow A=1$.
For $t=-4$,$-5 = -5B \Rightarrow B=1$.
Thus,$\frac{2{x^2} + 3}{{({x^2} - 1)({x^2} + 4)}} = \frac{1}{{({x^2} - 1)}} + \frac{1}{{({x^2} + 4)}}$.
$I = \int {\frac{{dx}}{{({x^2} - 1)}} + \int {\frac{{dx}}{{{x^2} + 2^2}}} }$.
Using standard integrals $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log|\frac{x-a}{x+a}|$ and $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$:
$I = \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + \frac{1}{2} \tan^{-1}(\frac{x}{2}) + c$.
Comparing with $a \log \left( \frac{x-1}{x+1} \right) + b \tan^{-1}(\frac{x}{2}) + c$,we get $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

(B) Let $I = \int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} dx$.
Substitute $x^{2} = y$.
Then,we can write the integrand as $\frac{y}{(y+1)(y+4)}$.
Using partial fractions,$\frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$.
$y = A(y+4) + B(y+1)$.
For $y = -1$,$-1 = A(3) \implies A = -\frac{1}{3}$.
For $y = -4$,$-4 = B(-3) \implies B = \frac{4}{3}$.
Thus,$I = \int \left( \frac{-1/3}{x^{2}+1} + \frac{4/3}{x^{2}+4} \right) dx$.
$I = -\frac{1}{3} \int \frac{1}{x^{2}+1} dx + \frac{4}{3} \int \frac{1}{x^{2}+2^{2}} dx$.
Using the formula $\int \frac{1}{x^{2}+a^{2}} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = -\frac{1}{3} \tan^{-1}(x) + \frac{4}{3} \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C$.
$I = -\frac{1}{3} \tan^{-1}(x) + \frac{2}{3} \tan^{-1}(\frac{x}{2}) + C$.

(A) The integrand is a proper rational function. We use the method of partial fractions to decompose it.
$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$ ..........$(1)$
Multiplying both sides by $(x+1)(x+2)$,we get:
$1 = A(x+2) + B(x+1)$
To find $A$,set $x = -1$:
$1 = A(-1+2) + B(0) \implies A = 1$
To find $B$,set $x = -2$:
$1 = A(0) + B(-2+1) \implies B = -1$
Substituting $A$ and $B$ back into equation $(1)$:
$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$
Now,integrating both sides:
$\int \frac{dx}{(x+1)(x+2)} = \int \frac{dx}{x+1} - \int \frac{dx}{x+2}$
$= \log |x+1| - \log |x+2| + C$
$= \log \left|\frac{x+1}{x+2}\right| + C$

(A) The integrand $\frac{x^{2}+1}{x^{2}-5 x+6}$ is an improper rational function. By performing long division,we get:
$\frac{x^{2}+1}{x^{2}-5 x+6} = 1 + \frac{5x-5}{x^{2}-5x+6} = 1 + \frac{5x-5}{(x-2)(x-3)}$
Using partial fractions for $\frac{5x-5}{(x-2)(x-3)}$:
$\frac{5x-5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$
$5x-5 = A(x-3) + B(x-2)$
Setting $x=2$: $5(2)-5 = A(2-3) \implies 5 = -A \implies A = -5$
Setting $x=3$: $5(3)-5 = B(3-2) \implies 10 = B$
Thus,$\frac{x^{2}+1}{x^{2}-5 x+6} = 1 - \frac{5}{x-2} + \frac{10}{x-3}$
Integrating both sides:
$\int \left( 1 - \frac{5}{x-2} + \frac{10}{x-3} \right) dx = \int dx - 5 \int \frac{1}{x-2} dx + 10 \int \frac{1}{x-3} dx$
$= x - 5 \log |x-2| + 10 \log |x-3| + C$

(N/A) The integrand is of the type as given in the table below:

Form of the rational function	Form of the partial fraction
$\frac{p x^{2}+q x+r}{(x-a)^{2}(x-b)}$	$\frac{A}{x-a}+\frac{B}{(x-a)^{2}}+\frac{C}{x-b}$

We write: $\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
Multiplying by the denominator,we get: $3 x-2 = A(x+1)(x+3)+B(x+3)+C(x+1)^{2}$
Expanding the terms: $3 x-2 = A(x^{2}+4 x+3)+B(x+3)+C(x^{2}+2 x+1)$
Comparing the coefficients of $x^{2}, x$ and the constant term on both sides,we get:
$A+C=0$
$4 A+B+2 C=3$
$3 A+3 B+C=-2$
Solving these equations,we get $A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$.
Thus,the integrand is: $\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Integrating both sides:
$\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x = \frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
$= \frac{11}{4} \log |x+1| + \frac{5}{2(x+1)} - \frac{11}{4} \log |x+3| + C$
$= \frac{11}{4} \log \left|\frac{x+1}{x+3}\right| + \frac{5}{2(x+1)} + C$

Consider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put $x^{2}=y$.
Then $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$.
Using partial fractions,write $\frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$.
This implies $y = A(y+4) + B(y+1)$.
Comparing coefficients of $y$ and constant terms,we get $A+B=1$ and $4A+B=0$.
Solving these,we find $A = -\frac{1}{3}$ and $B = \frac{4}{3}$.
Thus,$\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} = -\frac{1}{3(x^{2}+1)} + \frac{4}{3(x^{2}+4)}$.
Integrating both sides,$\int \frac{x^{2} dx}{(x^{2}+1)(x^{2}+4)} = -\frac{1}{3} \int \frac{dx}{x^{2}+1} + \frac{4}{3} \int \frac{dx}{x^{2}+4}$.
Using the standard integral $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$= -\frac{1}{3} \tan^{-1}(x) + \frac{4}{3} \times \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C$.
$= -\frac{1}{3} \tan^{-1}(x) + \frac{2}{3} \tan^{-1}(\frac{x}{2}) + C$.

(N/A) The integrand is a proper rational function. We decompose it into partial fractions as follows:

Form of the rational function	Form of the partial fraction
$\frac{px^{2}+qx+r}{(x-a)(x^{2}+bx+c)}$	$\frac{A}{x-a}+\frac{Bx+C}{x^{2}+bx+c}$

$\frac{x^{2}+x+1}{(x+2)(x^{2}+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^{2}+1}$
Multiplying both sides by $(x+2)(x^{2}+1)$,we get:
$x^{2}+x+1 = A(x^{2}+1) + (Bx+C)(x+2)$
$x^{2}+x+1 = A(x^{2}+1) + Bx^{2} + 2Bx + Cx + 2C$
$x^{2}+x+1 = (A+B)x^{2} + (2B+C)x + (A+2C)$
Equating the coefficients of $x^{2}, x$ and the constant term,we get:
$A+B = 1$
$2B+C = 1$
$A+2C = 1$
Solving these equations:
From $A+B=1$,$B=1-A$.
From $A+2C=1$,$C=\frac{1-A}{2}$.
Substituting into $2B+C=1$:
$2(1-A) + \frac{1-A}{2} = 1$
$4-4A+1-A = 2$
$5-5A = 2 \implies 5A = 3 \implies A = \frac{3}{5}$.
Then $B = 1 - \frac{3}{5} = \frac{2}{5}$ and $C = \frac{1 - 3/5}{2} = \frac{2/5}{2} = \frac{1}{5}$.
Thus,$\int \frac{x^{2}+x+1}{(x+2)(x^{2}+1)} dx = \int \left( \frac{3/5}{x+2} + \frac{2/5x + 1/5}{x^{2}+1} \right) dx$
$= \frac{3}{5} \int \frac{1}{x+2} dx + \frac{1}{5} \int \frac{2x}{x^{2}+1} dx + \frac{1}{5} \int \frac{1}{x^{2}+1} dx$
$= \frac{3}{5} \ln|x+2| + \frac{1}{5} \ln(x^{2}+1) + \frac{1}{5} \tan^{-1}(x) + C$.

(C) Let $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.
Multiplying both sides by $(x+1)(x+2)$,we get $x = A(x+2) + B(x+1)$.
To find $A$,put $x = -1$: $-1 = A(-1+2) \Rightarrow A = -1$.
To find $B$,put $x = -2$: $-2 = B(-2+1) \Rightarrow -2 = -B \Rightarrow B = 2$.
Thus,$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$.
Integrating both sides with respect to $x$:
$\int \frac{x}{(x+1)(x+2)} dx = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx$.
$= -\log |x+1| + 2\log |x+2| + C$.
$= \log |x+2|^2 - \log |x+1| + C$.
$= \log \left| \frac{(x+2)^2}{x+1} \right| + C$.

Let $\frac{1}{x^{2}-9} = \frac{1}{(x+3)(x-3)}$.
Using partial fractions,we write $\frac{1}{(x+3)(x-3)} = \frac{A}{x+3} + \frac{B}{x-3}$.
Multiplying both sides by $(x+3)(x-3)$,we get $1 = A(x-3) + B(x+3)$.
Setting $x = 3$,we get $1 = B(6)$,so $B = \frac{1}{6}$.
Setting $x = -3$,we get $1 = A(-6)$,so $A = -\frac{1}{6}$.
Thus,$\int \frac{1}{x^{2}-9} dx = \int \left( \frac{-1/6}{x+3} + \frac{1/6}{x-3} \right) dx$.
$= -\frac{1}{6} \int \frac{1}{x+3} dx + \frac{1}{6} \int \frac{1}{x-3} dx$.
$= -\frac{1}{6} \ln |x+3| + \frac{1}{6} \ln |x-3| + C$.
$= \frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C$,where $C$ is the constant of integration.

(A) Let $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.
$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.
For $x=1$: $3(1)-1 = A(1-2)(1-3) \Rightarrow 2 = A(-1)(-2) \Rightarrow 2A = 2 \Rightarrow A=1$.
For $x=2$: $3(2)-1 = B(2-1)(2-3) \Rightarrow 5 = B(1)(-1) \Rightarrow -B = 5 \Rightarrow B=-5$.
For $x=3$: $3(3)-1 = C(3-1)(3-2) \Rightarrow 8 = C(2)(1) \Rightarrow 2C = 8 \Rightarrow C=4$.
Thus,$\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}$.
Integrating both sides with respect to $x$:
$\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx$.
$= \log |x-1| - 5 \log |x-2| + 4 \log |x-3| + C$.

Let $\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.
Multiplying both sides by $(x-1)(x-2)(x-3)$,we get:
$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.
To find $A$,put $x=1$:
$1 = A(1-2)(1-3) \Rightarrow 1 = A(-1)(-2) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}$.
To find $B$,put $x=2$:
$2 = B(2-1)(2-3) \Rightarrow 2 = B(1)(-1) \Rightarrow 2 = -B \Rightarrow B = -2$.
To find $C$,put $x=3$:
$3 = C(3-1)(3-2) \Rightarrow 3 = C(2)(1) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}$.
Thus,$\frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} - \frac{2}{x-2} + \frac{3}{2(x-3)}$.
Integrating both sides:
$\int \frac{x}{(x-1)(x-2)(x-3)} dx = \frac{1}{2} \int \frac{1}{x-1} dx - 2 \int \frac{1}{x-2} dx + \frac{3}{2} \int \frac{1}{x-3} dx$.
$= \frac{1}{2} \log |x-1| - 2 \log |x-2| + \frac{3}{2} \log |x-3| + K$.

Let $\frac{2x}{x^{2}+3x+2} = \frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.
Multiplying both sides by $(x+1)(x+2)$,we get:
$2x = A(x+2) + B(x+1)$ ........$(1)$
To find $A$,put $x = -1$ in equation $(1)$:
$2(-1) = A(-1+2) + B(0)$
$-2 = A(1) \implies A = -2$.
To find $B$,put $x = -2$ in equation $(1)$:
$2(-2) = A(0) + B(-2+1)$
$-4 = B(-1) \implies B = 4$.
Therefore,$\frac{2x}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{4}{x+2}$.
Integrating both sides with respect to $x$:
$\int \frac{2x}{x^{2}+3x+2} dx = \int \left( \frac{4}{x+2} - \frac{2}{x+1} \right) dx$
$= 4 \int \frac{1}{x+2} dx - 2 \int \frac{1}{x+1} dx$
$= 4 \log |x+2| - 2 \log |x+1| + C$,where $C$ is an arbitrary constant.

The given integrand is $\frac{1-x^{2}}{x-2x^{2}}$. Since the degree of the numerator is equal to the degree of the denominator,it is an improper fraction.
First,we perform long division:
$\frac{1-x^{2}}{-2x^{2}+x} = \frac{1}{2} + \frac{2-x}{2x(1-2x)}$
Now,we decompose $\frac{2-x}{x(1-2x)}$ into partial fractions:
$\frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}$
$2-x = A(1-2x) + Bx$
Setting $x=0$,we get $A=2$.
Setting $x=\frac{1}{2}$,we get $2-\frac{1}{2} = B(\frac{1}{2}) \Rightarrow \frac{3}{2} = \frac{B}{2} \Rightarrow B=3$.
Thus,$\frac{1-x^{2}}{x(1-2x)} = \frac{1}{2} + \frac{1}{2} \left( \frac{2}{x} + \frac{3}{1-2x} \right) = \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)}$.
Integrating with respect to $x$:
$\int \left( \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)} \right) dx = \frac{x}{2} + \log|x| + \frac{3}{2} \cdot \frac{\log|1-2x|}{-2} + C$
$= \frac{x}{2} + \log|x| - \frac{3}{4} \log|1-2x| + C$.

Let $\frac{x}{(x^{2}+1)(x-1)} = \frac{Ax+B}{x^{2}+1} + \frac{C}{x-1}$
$x = (Ax+B)(x-1) + C(x^{2}+1)$
$x = Ax^{2} - Ax + Bx - B + Cx^{2} + C$
Equating the coefficients of $x^{2}$,$x$,and the constant term,we obtain:
$A+C = 0$
$-A+B = 1$
$-B+C = 0$
Solving these equations,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.
Substituting these values:
$\frac{x}{(x^{2}+1)(x-1)} = \frac{-\frac{1}{2}x + \frac{1}{2}}{x^{2}+1} + \frac{\frac{1}{2}}{x-1}$
Integrating both sides:
$\int \frac{x}{(x^{2}+1)(x-1)} dx = -\frac{1}{2} \int \frac{x}{x^{2}+1} dx + \frac{1}{2} \int \frac{1}{x^{2}+1} dx + \frac{1}{2} \int \frac{1}{x-1} dx$
$= -\frac{1}{4} \int \frac{2x}{x^{2}+1} dx + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \log|x-1| + C$
$= -\frac{1}{4} \log(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \log|x-1| + C$
$= \frac{1}{2} \log|x-1| - \frac{1}{4} \log(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) + C$

Let $\frac{x}{(x-1)^{2}(x+2)} = \frac{A}{(x-1)} + \frac{B}{(x-1)^{2}} + \frac{C}{(x+2)}$.
Multiplying both sides by $(x-1)^{2}(x+2)$,we get:
$x = A(x-1)(x+2) + B(x+2) + C(x-1)^{2}$.
For $x=1$,$1 = B(1+2) \Rightarrow 3B = 1 \Rightarrow B = \frac{1}{3}$.
For $x=-2$,$-2 = C(-2-1)^{2} \Rightarrow 9C = -2 \Rightarrow C = -\frac{2}{9}$.
Comparing the coefficients of $x^{2}$ on both sides: $0 = A + C \Rightarrow A = -C = \frac{2}{9}$.
Thus,$\frac{x}{(x-1)^{2}(x+2)} = \frac{2}{9(x-1)} + \frac{1}{3(x-1)^{2}} - \frac{2}{9(x+2)}$.
Integrating both sides:
$\int \frac{x}{(x-1)^{2}(x+2)} dx = \frac{2}{9} \int \frac{1}{x-1} dx + \frac{1}{3} \int (x-1)^{-2} dx - \frac{2}{9} \int \frac{1}{x+2} dx$.
$= \frac{2}{9} \log |x-1| + \frac{1}{3} \left( \frac{(x-1)^{-1}}{-1} \right) - \frac{2}{9} \log |x+2| + K$.
$= \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + K$,where $K$ is the constant of integration.

(A) First,factor the denominator: $x^{3}-x^{2}-x+1 = x^{2}(x-1)-1(x-1) = (x^{2}-1)(x-1) = (x-1)(x+1)(x-1) = (x-1)^{2}(x+1)$.
Using partial fractions,let $\frac{3x+5}{(x-1)^{2}(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{x+1}$.
Multiplying by the denominator,we get $3x+5 = A(x-1)(x+1) + B(x+1) + C(x-1)^{2}$.
Setting $x=1$,we get $3(1)+5 = B(1+1) \Rightarrow 8 = 2B \Rightarrow B=4$.
Setting $x=-1$,we get $3(-1)+5 = C(-1-1)^{2} \Rightarrow 2 = 4C \Rightarrow C=\frac{1}{2}$.
Comparing coefficients of $x^{2}$,we get $A+C=0 \Rightarrow A = -C = -\frac{1}{2}$.
Thus,$\frac{3x+5}{(x-1)^{2}(x+1)} = -\frac{1}{2(x-1)} + \frac{4}{(x-1)^{2}} + \frac{1}{2(x+1)}$.
Integrating both sides: $\int \frac{3x+5}{(x-1)^{2}(x+1)} dx = -\frac{1}{2} \int \frac{1}{x-1} dx + 4 \int (x-1)^{-2} dx + \frac{1}{2} \int \frac{1}{x+1} dx$.
$= -\frac{1}{2} \ln|x-1| - \frac{4}{x-1} + \frac{1}{2} \ln|x+1| + K$.
$= \frac{1}{2} \ln \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + K$,where $K$ is the constant of integration.

(N/A) We have $\frac{2x-3}{(x^2-1)(2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$.
Let $\frac{2x-3}{(x+1)(x-1)(2x+3)} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{2x+3}$.
Then $2x-3 = A(x-1)(2x+3) + B(x+1)(2x+3) + C(x+1)(x-1)$.
Setting $x=1$: $2(1)-3 = B(1+1)(2(1)+3) \Rightarrow -1 = B(2)(5) \Rightarrow B = -\frac{1}{10}$.
Setting $x=-1$: $2(-1)-3 = A(-1-1)(2(-1)+3) \Rightarrow -5 = A(-2)(1) \Rightarrow A = \frac{5}{2}$.
Setting $x=-\frac{3}{2}$: $2(-\frac{3}{2})-3 = C(-\frac{3}{2}+1)(-\frac{3}{2}-1) \Rightarrow -6 = C(-\frac{1}{2})(-\frac{5}{2}) \Rightarrow -6 = C(\frac{5}{4}) \Rightarrow C = -\frac{24}{5}$.
Thus,$\int \frac{2x-3}{(x^2-1)(2x+3)} dx = \int \left( \frac{5}{2(x+1)} - \frac{1}{10(x-1)} - \frac{24}{5(2x+3)} \right) dx$.
$= \frac{5}{2} \log|x+1| - \frac{1}{10} \log|x-1| - \frac{24}{5} \cdot \frac{1}{2} \log|2x+3| + K$.
$= \frac{5}{2} \log|x+1| - \frac{1}{10} \log|x-1| - \frac{12}{5} \log|2x+3| + K$,where $K$ is the constant of integration.

(N/A) Given the integral: $I = \int \frac{5x}{(x+1)(x^2-4)} dx$.
First,factor the denominator: $(x+1)(x^2-4) = (x+1)(x+2)(x-2)$.
Using partial fractions,let $\frac{5x}{(x+1)(x+2)(x-2)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x-2}$.
Then,$5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2)$.
Setting $x = -1$: $5(-1) = A(1)(-3) \Rightarrow -5 = -3A \Rightarrow A = \frac{5}{3}$.
Setting $x = -2$: $5(-2) = B(-1)(-4) \Rightarrow -10 = 4B \Rightarrow B = -\frac{5}{2}$.
Setting $x = 2$: $5(2) = C(3)(4) \Rightarrow 10 = 12C \Rightarrow C = \frac{5}{6}$.
Thus,$I = \int \left( \frac{5/3}{x+1} - \frac{5/2}{x+2} + \frac{5/6}{x-2} \right) dx$.
Integrating term by term,we get $I = \frac{5}{3} \ln|x+1| - \frac{5}{2} \ln|x+2| + \frac{5}{6} \ln|x-2| + K$,where $K$ is the constant of integration.

(N/A) The given integrand is an improper fraction because the degree of the numerator is greater than the degree of the denominator.
First,divide $(x^{3}+x+1)$ by $(x^{2}-1)$:
$\frac{x^{3}+x+1}{x^{2}-1} = x + \frac{2x+1}{x^{2}-1}$
Now,decompose $\frac{2x+1}{x^{2}-1}$ into partial fractions:
$\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
$2x+1 = A(x+1) + B(x-1)$
Setting $x=1$,we get $3 = 2A \Rightarrow A = \frac{3}{2}$.
Setting $x=-1$,we get $-1 = -2B \Rightarrow B = \frac{1}{2}$.
Thus,$\frac{x^{3}+x+1}{x^{2}-1} = x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)}$.
Integrating both sides:
$\int \left( x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)} \right) dx = \frac{x^{2}}{2} + \frac{3}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C$.

(N/A) Let $\frac{2}{(1-x)(1+x^{2})} = \frac{A}{1-x} + \frac{Bx+C}{1+x^{2}}$
Multiplying both sides by $(1-x)(1+x^{2})$,we get:
$2 = A(1+x^{2}) + (Bx+C)(1-x)$
$2 = A + Ax^{2} + Bx - Bx^{2} + C - Cx$
$2 = (A-B)x^{2} + (B-C)x + (A+C)$
Equating the coefficients of $x^{2}, x,$ and the constant term,we obtain:
$A-B = 0 \Rightarrow A = B$
$B-C = 0 \Rightarrow B = C$
$A+C = 2$
Substituting $A=B$ and $C=B$ into $A+C=2$,we get $B+B=2$,so $B=1$. Thus,$A=1$ and $C=1$.
Therefore,$\frac{2}{(1-x)(1+x^{2})} = \frac{1}{1-x} + \frac{x+1}{1+x^{2}}$
Integrating both sides:
$\int \frac{2}{(1-x)(1+x^{2})} dx = \int \frac{1}{1-x} dx + \int \frac{x}{1+x^{2}} dx + \int \frac{1}{1+x^{2}} dx$
$= -\int \frac{-1}{1-x} dx + \frac{1}{2} \int \frac{2x}{1+x^{2}} dx + \int \frac{1}{1+x^{2}} dx$
$= -\log|1-x| + \frac{1}{2} \log|1+x^{2}| + \tan^{-1}(x) + K$
Where $K$ is the constant of integration.

Let $\frac{3x-1}{(x+2)^{2}} = \frac{A}{(x+2)} + \frac{B}{(x+2)^{2}}$
Multiplying both sides by $(x+2)^{2}$,we get:
$3x-1 = A(x+2) + B$
Equating the coefficients of $x$ and the constant terms:
$A = 3$
$2A + B = -1$
Substituting $A = 3$ into the second equation:
$2(3) + B = -1 \Rightarrow 6 + B = -1 \Rightarrow B = -7$
Thus,$\frac{3x-1}{(x+2)^{2}} = \frac{3}{(x+2)} - \frac{7}{(x+2)^{2}}$
Integrating both sides with respect to $x$:
$\int \frac{3x-1}{(x+2)^{2}} dx = 3 \int \frac{1}{x+2} dx - 7 \int (x+2)^{-2} dx$
$= 3 \log |x+2| - 7 \left( \frac{(x+2)^{-1}}{-1} \right) + C$
$= 3 \log |x+2| + \frac{7}{x+2} + C$
Where $C$ is the constant of integration.

(N/A) We have $\frac{1}{x^{4}-1} = \frac{1}{(x^{2}-1)(x^{2}+1)} = \frac{1}{(x+1)(x-1)(x^{2}+1)}$.
Using partial fractions,let $\frac{1}{(x+1)(x-1)(x^{2}+1)} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{Cx+D}{x^{2}+1}$.
Multiplying by $(x+1)(x-1)(x^{2}+1)$,we get $1 = A(x-1)(x^{2}+1) + B(x+1)(x^{2}+1) + (Cx+D)(x^{2}-1)$.
Comparing coefficients of $x^{3}, x^{2}, x,$ and the constant term:
$A+B+C = 0$
$-A+B+D = 0$
$A+B-C = 0$
$-A+B-D = 1$
Solving these equations,we get $A = -\frac{1}{4}, B = \frac{1}{4}, C = 0, D = -\frac{1}{2}$.
Thus,$\frac{1}{x^{4}-1} = -\frac{1}{4(x+1)} + \frac{1}{4(x-1)} - \frac{1}{2(x^{2}+1)}$.
Integrating both sides,$\int \frac{1}{x^{4}-1} dx = -\frac{1}{4} \int \frac{1}{x+1} dx + \frac{1}{4} \int \frac{1}{x-1} dx - \frac{1}{2} \int \frac{1}{x^{2}+1} dx$.
$= -\frac{1}{4} \log |x+1| + \frac{1}{4} \log |x-1| - \frac{1}{2} \tan^{-1} x + C$.
$= \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \tan^{-1} x + C$,where $C$ is the constant of integration.