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

To integrate $\int \frac{1}{x(x^{n}+1)} dx$,we multiply the numerator and denominator by $x^{n-1}$:
$\int \frac{1}{x(x^{n}+1)} dx = \int \frac{x^{n-1}}{x^{n}(x^{n}+1)} dx$
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:
$\int \frac{1}{n} \frac{dt}{t(t+1)} = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$
Integrating the terms:
$= \frac{1}{n} [\log |t| - \log |t+1|] + C$
$= \frac{1}{n} \log \left| \frac{t}{t+1} \right| + C$
Substituting $t = x^{n}$ back:
$= \frac{1}{n} \log \left| \frac{x^{n}}{x^{n}+1} \right| + C$,where $C$ is an arbitrary constant.

(D) Let $I = \int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx$.
Put $\sin x = t$,then $\cos x dx = dt$.
The integral becomes $I = \int \frac{dt}{(1-t)(2-t)}$.
Using partial fractions: $\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}$.
$1 = A(2-t) + B(1-t)$.
For $t = 1$,$1 = A(2-1) \Rightarrow A = 1$.
For $t = 2$,$1 = B(1-2) \Rightarrow B = -1$.
Thus,$\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}$.
Integrating both sides: $I = \int \left( \frac{1}{1-t} - \frac{1}{2-t} \right) dt$.
$I = -\ln|1-t| - (-\ln|2-t|) + C = \ln|2-t| - \ln|1-t| + C$.
$I = \ln\left|\frac{2-t}{1-t}\right| + C$.
Substituting $t = \sin x$,we get $I = \ln\left|\frac{2-\sin x}{1-\sin x}\right| + C$.

Given the integrand: $f(x) = \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}$
First,expand the numerator and denominator: $f(x) = \frac{x^{4}+3x^{2}+2}{x^{4}+7x^{2}+12}$
Perform polynomial division: $f(x) = 1 - \frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)}$
Using partial fractions for the second term: $\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)} = \frac{A}{x^{2}+3} + \frac{B}{x^{2}+4}$
Let $y = x^{2}$. Then $\frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4}$
$4y+10 = A(y+4) + B(y+3)$
For $y = -3$: $4(-3)+10 = A(-3+4) \Rightarrow -2 = A$
For $y = -4$: $4(-4)+10 = B(-4+3) \Rightarrow -6 = -B \Rightarrow B = 6$
Thus,$\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)} = \frac{-2}{x^{2}+3} + \frac{6}{x^{2}+4}$
Substituting back: $f(x) = 1 - (\frac{-2}{x^{2}+3} + \frac{6}{x^{2}+4}) = 1 + \frac{2}{x^{2}+3} - \frac{6}{x^{2}+4}$
Integrating: $\int f(x) dx = \int (1 + \frac{2}{x^{2}+(\sqrt{3})^{2}} - \frac{6}{x^{2}+2^{2}}) dx$
$= x + 2(\frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}}) - 6(\frac{1}{2} \tan^{-1} \frac{x}{2}) + C$
$= x + \frac{2}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} - 3 \tan^{-1} \frac{x}{2} + C$

(A) Let $I = \int \frac{2x}{(x^2+1)(x^2+3)} dx$.
Substitute $x^2 = t$,then $2x dx = dt$.
Thus,the integral becomes $I = \int \frac{dt}{(t+1)(t+3)}$.
Using partial fractions,let $\frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3}$.
This implies $1 = A(t+3) + B(t+1)$.
Setting $t = -1$,we get $1 = A(2) \Rightarrow A = \frac{1}{2}$.
Setting $t = -3$,we get $1 = B(-2) \Rightarrow B = -\frac{1}{2}$.
Therefore,$I = \int \left( \frac{1}{2(t+1)} - \frac{1}{2(t+3)} \right) dt$.
Integrating,we get $I = \frac{1}{2} \ln|t+1| - \frac{1}{2} \ln|t+3| + C$.
Substituting $t = x^2$ back,we get $I = \frac{1}{2} \ln \left| \frac{x^2+1}{x^2+3} \right| + C$,where $C$ is the constant of integration.

(N/A) To integrate $\int \frac{1}{x(x^{4}-1)} dx$,multiply the numerator and denominator by $x^{3}$:
$\int \frac{x^{3}}{x^{4}(x^{4}-1)} dx$
Let $x^{4} = t$,then $4x^{3} dx = dt$,which implies $x^{3} dx = \frac{1}{4} dt$.
Substituting these into the integral:
$\frac{1}{4} \int \frac{dt}{t(t-1)}$
Using partial fractions,$\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1}$.
$1 = A(t-1) + Bt$. Setting $t=0$,we get $A=-1$. Setting $t=1$,we get $B=1$.
Thus,$\frac{1}{4} \int (\frac{-1}{t} + \frac{1}{t-1}) dt = \frac{1}{4} [-\ln|t| + \ln|t-1|] + C$
$= \frac{1}{4} \ln|\frac{t-1}{t}| + C$
$= \frac{1}{4} \ln|\frac{x^{4}-1}{x^{4}}| + C$,where $C$ is the constant of integration.

Let $e^x = t$. Then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.
Substituting these into the integral,we get:
$\int \frac{1}{e^x - 1} dx = \int \frac{1}{t - 1} \cdot \frac{dt}{t} = \int \frac{1}{t(t - 1)} dt$.
Using partial fractions:
$\frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1}$.
$1 = A(t - 1) + Bt$.
Setting $t = 0$,we get $A = -1$.
Setting $t = 1$,we get $B = 1$.
Thus,$\int \frac{1}{t(t - 1)} dt = \int \left( \frac{-1}{t} + \frac{1}{t - 1} \right) dt$.
$= -\log|t| + \log|t - 1| + C = \log\left| \frac{t - 1}{t} \right| + C$.
Substituting $t = e^x$ back:
$= \log\left| \frac{e^x - 1}{e^x} \right| + C$.

(D) Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
By partial fraction decomposition,$x = A(x-2) + B(x-1)$.
Setting $x = 1$,we get $1 = A(1-2) \Rightarrow A = -1$.
Setting $x = 2$,we get $2 = B(2-1) \Rightarrow B = 2$.
Thus,$\frac{x}{(x-1)(x-2)} = -\frac{1}{x-1} + \frac{2}{x-2}$.
Integrating both sides with respect to $x$:
$\int \frac{x \, dx}{(x-1)(x-2)} = \int \left( -\frac{1}{x-1} + \frac{2}{x-2} \right) dx$.
$= -\log |x-1| + 2 \log |x-2| + C$.
$= \log |x-2|^2 - \log |x-1| + C$.
$= \log \left| \frac{(x-2)^2}{x-1} \right| + C$.
Therefore,the correct option is $D$.

(A) To evaluate the integral $\int \frac{dx}{x(x^{2}+1)}$,we use partial fractions.
Let $\frac{1}{x(x^{2}+1)} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1}$.
Multiplying both sides by $x(x^{2}+1)$,we get $1 = A(x^{2}+1) + (Bx+C)x$.
Comparing coefficients of $x^{2}$,$x$,and the constant term,we get $A+B=0$,$C=0$,and $A=1$.
Thus,$A=1$,$B=-1$,and $C=0$.
So,$\frac{1}{x(x^{2}+1)} = \frac{1}{x} - \frac{x}{x^{2}+1}$.
Integrating both sides,we get $\int \frac{dx}{x} - \int \frac{x}{x^{2}+1} dx$.
$= \log |x| - \frac{1}{2} \log |x^{2}+1| + C$.
Since $x^{2}+1 > 0$ for all real $x$,we can write this as $\log |x| - \frac{1}{2} \log (x^{2}+1) + C$.

(A) We perform polynomial division to simplify the integrand:
$\frac{x^{4}}{(x-1)(x^{2}+1)} = \frac{x^{4}}{x^{3}-x^{2}+x-1} = (x+1) + \frac{1}{x^{3}-x^{2}+x-1} = (x+1) + \frac{1}{(x-1)(x^{2}+1)}$
Using partial fractions for $\frac{1}{(x-1)(x^{2}+1)}$:
$\frac{1}{(x-1)(x^{2}+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^{2}+1}$
$1 = A(x^{2}+1) + (Bx+C)(x-1) = (A+B)x^{2} + (C-B)x + (A-C)$
Equating coefficients: $A+B=0$,$C-B=0$,$A-C=1$. Solving these gives $A=\frac{1}{2}$,$B=-\frac{1}{2}$,$C=-\frac{1}{2}$.
Substituting back:
$\int \frac{x^{4} dx}{(x-1)(x^{2}+1)} = \int (x+1) dx + \frac{1}{2} \int \frac{dx}{x-1} - \frac{1}{2} \int \frac{x dx}{x^{2}+1} - \frac{1}{2} \int \frac{dx}{x^{2}+1}$
$= \frac{x^{2}}{2} + x + \frac{1}{2} \ln|x-1| - \frac{1}{4} \ln(x^{2}+1) - \frac{1}{2} \tan^{-1}(x) + C$

(D) We have $\frac{1}{x-x^{3}} = \frac{1}{x(1-x^{2})} = \frac{1}{x(1-x)(1+x)}$.
Using partial fractions,let $\frac{1}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} \dots (1)$.
Multiplying by $x(1-x)(1+x)$,we get $1 = A(1-x^{2}) + Bx(1+x) + Cx(1-x)$.
Setting $x=0$,we get $A=1$.
Setting $x=1$,we get $1 = B(1)(2) \Rightarrow B = \frac{1}{2}$.
Setting $x=-1$,we get $1 = C(-1)(2) \Rightarrow C = -\frac{1}{2}$.
Substituting these values into $(1)$,we have $\frac{1}{x-x^{3}} = \frac{1}{x} + \frac{1}{2(1-x)} - \frac{1}{2(1+x)}$.
Integrating both sides with respect to $x$:
$\int \frac{1}{x-x^{3}} dx = \int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{1-x} dx - \frac{1}{2} \int \frac{1}{1+x} dx$.
$= \log |x| - \frac{1}{2} \log |1-x| - \frac{1}{2} \log |1+x| + C$.
$= \log |x| - \frac{1}{2} \log |(1-x)(1+x)| + C$.
$= \log |x| - \frac{1}{2} \log |1-x^{2}| + C$.
$= \frac{1}{2} \log |x^{2}| - \frac{1}{2} \log |1-x^{2}| + C$.
$= \frac{1}{2} \log \left| \frac{x^{2}}{1-x^{2}} \right| + C$.

Let $\frac{5 x}{(x+1)(x^{2}+9)} = \frac{A}{(x+1)} + \frac{Bx+C}{(x^{2}+9)}$ ........$(1)$
$\Rightarrow 5x = A(x^{2}+9) + (Bx+C)(x+1)$
$\Rightarrow 5x = Ax^{2} + 9A + Bx^{2} + Bx + Cx + C$
Equating the coefficients of $x^{2}, x,$ and the constant term,we obtain:
$A+B = 0$
$B+C = 5$
$9A+C = 0$
On solving these equations,we obtain $A = -\frac{1}{2}, B = \frac{1}{2},$ and $C = \frac{9}{2}$.
From equation $(1)$,we have:
$\frac{5x}{(x+1)(x^{2}+9)} = -\frac{1}{2(x+1)} + \frac{\frac{x}{2} + \frac{9}{2}}{(x^{2}+9)}$
$\int \frac{5x}{(x+1)(x^{2}+9)} dx = \int \left\{ -\frac{1}{2(x+1)} + \frac{x+9}{2(x^{2}+9)} \right\} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{2} \int \frac{x}{x^{2}+9} dx + \frac{9}{2} \int \frac{1}{x^{2}+9} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \int \frac{2x}{x^{2}+9} dx + \frac{9}{2} \int \frac{1}{x^{2}+9} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \log (x^{2}+9) + \frac{9}{2} \cdot \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \log (x^{2}+9) + \frac{3}{2} \tan^{-1} \left( \frac{x}{3} \right) + C$

(N/A) Let $I = \int \frac{e^{x}}{(1+e^{x})(2+e^{x})} dx$.
Substitute $e^{x} = t$,which implies $e^{x} dx = dt$.
The integral becomes $\int \frac{dt}{(t+1)(t+2)}$.
Using partial fractions,we write $\frac{1}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$.
Solving for constants,$1 = A(t+2) + B(t+1)$.
For $t = -1$,$A = 1$. For $t = -2$,$B = -1$.
Thus,$\int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) dt = \log|t+1| - \log|t+2| + C$.
Using the property $\log a - \log b = \log(\frac{a}{b})$,we get $\log|\frac{t+1}{t+2}| + C$.
Substituting back $t = e^{x}$,the final result is $\log|\frac{1+e^{x}}{2+e^{x}}| + C$.

Let $\frac{1}{(x^{2}+1)(x^{2}+4)} = \frac{A}{x^{2}+1} + \frac{B}{x^{2}+4}$.
Substituting $y = x^{2}$,we have $\frac{1}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$.
$1 = A(y+4) + B(y+1)$.
For $y = -1$,$1 = A(3) \Rightarrow A = \frac{1}{3}$.
For $y = -4$,$1 = B(-3) \Rightarrow B = -\frac{1}{3}$.
Thus,$\frac{1}{(x^{2}+1)(x^{2}+4)} = \frac{1}{3(x^{2}+1)} - \frac{1}{3(x^{2}+4)}$.
Integrating both sides:
$\int \frac{1}{(x^{2}+1)(x^{2}+4)} dx = \frac{1}{3} \int \frac{1}{x^{2}+1} dx - \frac{1}{3} \int \frac{1}{x^{2}+2^{2}} dx$.
Using the standard integrals $\int \frac{1}{x^{2}+a^{2}} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$= \frac{1}{3} \tan^{-1}(x) - \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C$.
$= \frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}(\frac{x}{2}) + C$.

(N/A) Let $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^{2}} + \frac{C}{x+2}$ ..........$(1)$
Multiplying both sides by $(x+1)^{2}(x+2)$,we get:
$x^{2}+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)^{2}$
Expanding the terms:
$x^{2}+x+1 = A(x^{2}+3x+2) + B(x+2) + C(x^{2}+2x+1)$
$x^{2}+x+1 = (A+C)x^{2} + (3A+B+2C)x + (2A+2B+C)$
Equating the coefficients of $x^{2}, x$ and the constant term:
$A+C = 1$
$3A+B+2C = 1$
$2A+2B+C = 1$
Solving these equations:
From $A+C=1$,we have $C = 1-A$.
Substituting into the other equations:
$3A+B+2(1-A) = 1 \Rightarrow A+B = -1 \Rightarrow B = -1-A$
$2A+2(-1-A)+(1-A) = 1 \Rightarrow 2A-2-2A+1-A = 1 \Rightarrow -A-1 = 1 \Rightarrow A = -2$
Then $B = -1-(-2) = 1$ and $C = 1-(-2) = 3$.
Substituting values into equation $(1)$:
$\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} dx = \int \left( \frac{-2}{x+1} + \frac{1}{(x+1)^{2}} + \frac{3}{x+2} \right) dx$
$= -2 \ln|x+1| - \frac{1}{x+1} + 3 \ln|x+2| + K$

(D) Given integral: $I = \int \frac{\cos \theta d \theta}{5+7 \sin \theta-2 \cos ^{2} \theta}$
Substituting $\cos^2 \theta = 1 - \sin^2 \theta$:
$I = \int \frac{\cos \theta d \theta}{5+7 \sin \theta-2(1-\sin^2 \theta)} = \int \frac{\cos \theta d \theta}{2 \sin^2 \theta+7 \sin \theta+3}$
Let $\sin \theta = t$,then $\cos \theta d \theta = dt$:
$I = \int \frac{dt}{2t^2+7t+3} = \int \frac{dt}{(2t+1)(t+3)}$
Using partial fractions:
$\frac{1}{(2t+1)(t+3)} = \frac{1}{5} \left( \frac{2}{2t+1} - \frac{1}{t+3} \right)$
Integrating:
$I = \frac{1}{5} \int \left( \frac{2}{2t+1} - \frac{1}{t+3} \right) dt = \frac{1}{5} \ln \left| \frac{2t+1}{t+3} \right| + C$
Substituting $t = \sin \theta$ back:
$I = \frac{1}{5} \ln \left| \frac{2 \sin \theta+1}{\sin \theta+3} \right| + C$
Comparing with $A \log _{e}|B(\theta)|+C$,we get $A = \frac{1}{5}$ and $B(\theta) = \frac{2 \sin \theta+1}{\sin \theta+3}$.
Therefore,$\frac{B(\theta)}{A} = \frac{\frac{2 \sin \theta+1}{\sin \theta+3}}{\frac{1}{5}} = \frac{5(2 \sin \theta+1)}{\sin \theta+3}$.

(D) Divide the numerator and denominator by $\cos^3 x$:
$I = \int \frac{\tan x \sec^2 x}{1+\tan^3 x} dx$
Let $\tan x = t$,then $\sec^2 x dx = dt$:
$I = \int \frac{t}{(t+1)(t^2-t+1)} dt$
Using partial fractions: $\frac{t}{(t+1)(t^2-t+1)} = \frac{A}{t+1} + \frac{Bt+C}{t^2-t+1}$
$t = A(t^2-t+1) + (Bt+C)(t+1)$
Setting $t = -1$: $-1 = A(1+1+1) \Rightarrow A = -\frac{1}{3}$
Comparing coefficients of $t^2$: $A+B = 0 \Rightarrow B = \frac{1}{3}$
Comparing constant terms: $A+C = 0 \Rightarrow C = \frac{1}{3}$
So,$I = -\frac{1}{3} \int \frac{dt}{t+1} + \int \frac{\frac{1}{3}t + \frac{1}{3}}{t^2-t+1} dt$
$I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \int \frac{2t-1+3}{t^2-t+1} dt$
$I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2-t+1| + \frac{1}{2} \int \frac{dt}{(t-1/2)^2 + 3/4}$
$I = -\frac{1}{3} \ln|1+\tan x| + \frac{1}{6} \ln|1-\tan x+\tan^2 x| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right) + C$
Thus,$\alpha = -\frac{1}{3}, \beta = \frac{1}{6}, \gamma = \frac{1}{\sqrt{3}}$
$18(\alpha+\beta+\gamma^2) = 18(-\frac{1}{3} + \frac{1}{6} + \frac{1}{3}) = 18(\frac{1}{6}) = 3$.

(A) Let $t = x^2$,then $dt = 2x dx$.
Substituting this into the integral,we get:
$f(x) = \int \frac{dt}{(t+1)(t+3)}$.
Using partial fractions:
$\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right)$.
Integrating both sides:
$f(x) = \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} \right) dt = \frac{1}{2} (\ln|t+1| - \ln|t+3|) + C$.
Substituting $t = x^2$ back:
$f(x) = \frac{1}{2} \ln \left( \frac{x^2+1}{x^2+3} \right) + C$.
Given $f(3) = \frac{1}{2}(\ln 10 - \ln 12) + C = \frac{1}{2}(\ln 5 - \ln 6) + C$.
Since $\frac{1}{2}(\ln 10 - \ln 12) = \frac{1}{2}(\ln(2 \times 5) - \ln(2 \times 6)) = \frac{1}{2}(\ln 5 - \ln 6)$,we find $C = 0$.
Thus,$f(x) = \frac{1}{2} \ln \left( \frac{x^2+1}{x^2+3} \right)$.
Calculating $f(4)$:
$f(4) = \frac{1}{2} \ln \left( \frac{4^2+1}{4^2+3} \right) = \frac{1}{2} \ln \left( \frac{17}{19} \right) = \frac{1}{2} (\ln 17 - \ln 19)$.

(D) Given $I(x) = \int \frac{x+1}{x(1+x e^x)^2} dx$.
Multiply numerator and denominator by $e^x$:
$I(x) = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} dx$.
Let $u = x e^x$,then $du = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting this into the integral:
$I(x) = \int \frac{du}{u(1+u)^2}$.
Using partial fractions: $\frac{1}{u(1+u)^2} = \frac{A}{u} + \frac{B}{1+u} + \frac{C}{(1+u)^2}$.
$1 = A(1+u)^2 + Bu(1+u) + Cu$.
For $u=0$,$A=1$. For $u=-1$,$C=-1$.
Comparing coefficients of $u^2$: $A+B=0 \implies B=-1$.
So,$I(x) = \int (\frac{1}{u} - \frac{1}{1+u} - \frac{1}{(1+u)^2}) du = \log|u| - \log|1+u| + \frac{1}{1+u} + C$.
$I(x) = \log|\frac{u}{1+u}| + \frac{1}{1+u} + C = \log|\frac{x e^x}{1+x e^x}| + \frac{1}{1+x e^x} + C$.
As $x \rightarrow \infty$,$\frac{x e^x}{1+x e^x} \rightarrow 1$,so $\log(1) = 0$ and $\frac{1}{1+x e^x} \rightarrow 0$.
Thus,$\lim_{x \rightarrow \infty} I(x) = 0 + 0 + C = 0 \implies C = 0$.
$I(1) = \log(\frac{e}{1+e}) + \frac{1}{1+e} = \log(e) - \log(1+e) + \frac{1}{1+e} = 1 - \log(1+e) + \frac{1}{1+e} = \frac{1+e+1}{1+e} - \log(1+e) = \frac{e+2}{e+1} - \log_e(e+1)$.

(B) Let $I = \int \limits_0^{\infty} \frac{6}{(e^x+1)(e^x+2)(e^x+3)} dx$.
Using partial fractions,we write:
$\frac{6}{(e^x+1)(e^x+2)(e^x+3)} = \frac{3}{e^x+1} - \frac{6}{e^x+2} + \frac{3}{e^x+3}$.
Substituting $u = e^x$,then $du = e^x dx$,so $dx = \frac{du}{u}$.
$I = \int_1^{\infty} \frac{6}{u(u+1)(u+2)(u+3)} du$.
Using partial fractions for the integrand:
$\frac{6}{u(u+1)(u+2)(u+3)} = \frac{1}{u} - \frac{3}{u+1} + \frac{3}{u+2} - \frac{1}{u+3}$.
Integrating term by term:
$I = \left[ \ln|u| - 3\ln|u+1| + 3\ln|u+2| - \ln|u+3| \right]_1^{\infty}$.
$I = \left[ \ln \left| \frac{u(u+2)^3}{(u+1)^3(u+3)} \right| \right]_1^{\infty}$.
As $u \to \infty$,the argument of the logarithm approaches $\ln(1) = 0$.
At $u = 1$,the value is $\ln \left( \frac{1(3)^3}{(2)^3(4)} \right) = \ln \left( \frac{27}{32} \right)$.
Therefore,$I = 0 - \ln \left( \frac{27}{32} \right) = \ln \left( \frac{32}{27} \right)$.

(D) To evaluate the integral $\int_1^2 \frac{x}{(x+2)(x+3)} \, dx$,we use partial fractions.
Let $\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$.
Multiplying by $(x+2)(x+3)$,we get $x = A(x+3) + B(x+2)$.
Setting $x = -2$,we get $-2 = A(1)$,so $A = -2$.
Setting $x = -3$,we get $-3 = B(-1)$,so $B = 3$.
Thus,$\int_1^2 \left( \frac{-2}{x+2} + \frac{3}{x+3} \right) \, dx = [-2 \log|x+2| + 3 \log|x+3|]_1^2$.
Evaluating at the limits:
$(-2 \log 4 + 3 \log 5) - (-2 \log 3 + 3 \log 4) = -5 \log 4 + 3 \log 5 + 2 \log 3 = \log(5^3 \cdot 3^2 / 4^5) = \log(125 \cdot 9 / 1024) = \log(\frac{1125}{1024})$.
Therefore,the correct option is $D$.

(A) Let $I = \int \frac{dx}{2e^{2x} + 3e^x + 1}$.
Put $e^x = t$,then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.
Substituting these into the integral,we get $I = \int \frac{dt}{t(2t^2 + 3t + 1)} = \int \frac{dt}{t(2t+1)(t+1)}$.
Using partial fractions: $\frac{1}{t(2t+1)(t+1)} = \frac{A}{t} + \frac{B}{2t+1} + \frac{C}{t+1}$.
$1 = A(2t+1)(t+1) + Bt(t+1) + Ct(2t+1)$.
For $t=0$,$1 = A(1)(1) \implies A=1$.
For $t=-1$,$1 = C(-1)(-2+1) = C(-1)(-1) = C \implies C=1$.
For $t=-1/2$,$1 = B(-1/2)(1/2) = -B/4 \implies B=-4$.
Thus,$I = \int (\frac{1}{t} - \frac{4}{2t+1} + \frac{1}{t+1}) dt$.
$I = \log|t| - 4 \cdot \frac{1}{2} \log|2t+1| + \log|t+1| + c$.
$I = \log|e^x| - 2 \log|2e^x+1| + \log|e^x+1| + c$.
$I = x + \log(e^x+1) - 2 \log(2e^x+1) + c$.

(A) To evaluate the integral $I = \int \frac{dx}{x(x^2+1)}$,we use partial fractions.
Let $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$.
Multiplying by $x(x^2+1)$,we get $1 = A(x^2+1) + (Bx+C)x$.
$1 = (A+B)x^2 + Cx + A$.
Comparing coefficients,we get $A=1$,$C=0$,and $A+B=0$,which implies $B=-1$.
Thus,$\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}$.
Integrating both sides,$I = \int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx$.
For the second integral,let $u = x^2+1$,then $du = 2x dx$,so $x dx = \frac{du}{2}$.
$I = \log |x| - \frac{1}{2} \int \frac{du}{u} = \log |x| - \frac{1}{2} \log |x^2+1| + c$.
Since $x^2+1 > 0$,we can write this as $\log |x| - \frac{1}{2} \log (x^2+1) + c$.

(A) Let $I = \int \frac{x^3}{x^4+5x^2+4} dx$.
Substitute $t = x^2$,then $dt = 2x dx$,or $x dx = \frac{1}{2} dt$.
The integral becomes $I = \frac{1}{2} \int \frac{t}{t^2+5t+4} dt$.
Factor the denominator: $t^2+5t+4 = (t+1)(t+4)$.
Using partial fractions: $\frac{t}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.
$t = A(t+4) + B(t+1)$.
For $t = -1$,$-1 = 3A \implies A = -1/3$.
For $t = -4$,$-4 = -3B \implies B = 4/3$.
So,$I = \frac{1}{2} \int \left( \frac{4/3}{t+4} - \frac{1/3}{t+1} \right) dt = \frac{1}{6} \int \left( \frac{4}{t+4} - \frac{1}{t+1} \right) dt$.
$I = \frac{1}{6} [4 \log|t+4| - \log|t+1|] + c$.
$I = \frac{1}{6} \log \left| \frac{(t+4)^4}{t+1} \right| + c$.
Substituting $t = x^2$ back: $I = \frac{1}{6} \log \left( \frac{(x^2+4)^4}{x^2+1} \right) + c = \frac{1}{3} \log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$.

(C) Let $I = \int \frac{dx}{x(x^3+1)}$.
Multiply the numerator and denominator by $x^2$:
$I = \int \frac{x^2 dx}{x^3(x^3+1)}$.
Let $x^3 = t$,then $3x^2 dx = dt$,or $x^2 dx = \frac{dt}{3}$.
Substituting these into the integral:
$I = \frac{1}{3} \int \frac{dt}{t(t+1)}$.
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
$I = \frac{1}{3} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt$.
$I = \frac{1}{3} (\log|t| - \log|t+1|) + c$.
$I = \frac{1}{3} \log \left|\frac{t}{t+1}\right| + c$.
Substituting $t = x^3$ back:
$I = \frac{1}{3} \log \left|\frac{x^3}{x^3+1}\right| + c = \log \left|\left(\frac{x^3}{x^3+1}\right)^{1/3}\right| + c = \log \left(\sqrt[3]{\frac{x^3}{x^3+1}}\right) + c$.

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,dx$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,dx$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(D) Let $I = \int \frac{x+1}{x(1+x e^x)^2} dx$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} dx$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting this into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants: $1 = A(1+t)^2 + Bt(1+t) + Ct$.
For $t=0$,$A=1$. For $t=-1$,$C=-1$. Comparing coefficients of $t^2$,$A+B=0 \Rightarrow B=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log |t| - \log |1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$:
$I = \log |x e^x| - \log |1+x e^x| + \frac{1}{1+x e^x} + c$.
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(B) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
Integrating term by term:
$I = \log |t| - \log |1+t| + \frac{1}{1+t} + C$.
$I = \log \left| \frac{t}{1+t} \right| + \frac{1}{1+t} + C$.
Substituting $t = x e^x$ back:
$I = \log \left( \frac{x e^x}{1+x e^x} \right) + \frac{1}{1+x e^x} + C$.

(C) Let $t = e^x$,then $dt = e^x dx$.
Substituting this into the integral,we get:
$\int \frac{dt}{(t+2)(t+1)}$.
Using partial fractions:
$\frac{1}{(t+2)(t+1)} = \frac{A}{t+2} + \frac{B}{t+1}$.
$1 = A(t+1) + B(t+2)$.
For $t = -1$,$B = 1$.
For $t = -2$,$A = -1$.
So,$\int (\frac{-1}{t+2} + \frac{1}{t+1}) dt = -\log|t+2| + \log|t+1| + C$.
$= \log|\frac{t+1}{t+2}| + C$.
Substituting $t = e^x$ back,we get:
$\log \left(\frac{e^x+1}{e^x+2}\right) + C$.

(D) Let $I = \int \frac{dx}{(\sin x + \cos x)(2 \cos x + \sin x)}$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{(\tan x + 1)(2 + \tan x)} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$.
$I = \int \frac{dt}{(t+1)(t+2)}$.
Using partial fractions: $\frac{1}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$.
$1 = A(t+2) + B(t+1)$.
For $t = -1$,$A = 1$. For $t = -2$,$B = -1$.
$I = \int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) dt = \log|t+1| - \log|t+2| + C$.
$I = \log \left| \frac{t+1}{t+2} \right| + C = \log \left| \frac{\tan x + 1}{\tan x + 2} \right| + C$.

(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9} \right)$.
Integrating both sides with respect to $x$:
$\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = \frac{1}{5} \int \frac{1}{x^{2} + 2^{2}} dx - \frac{1}{5} \int \frac{1}{x^{2} + 3^{2}} dx$.
Using the standard formula $\int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$:
$= \frac{1}{5} \left( \frac{1}{2} \tan^{-1} \frac{x}{2} \right) - \frac{1}{5} \left( \frac{1}{3} \tan^{-1} \frac{x}{3} \right) + C$.
$= \frac{1}{10} \tan^{-1} \frac{x}{2} - \frac{1}{15} \tan^{-1} \frac{x}{3} + C$.
Comparing this with $A \tan^{-1} \frac{x}{2} + B \tan^{-1} \frac{x}{3} + C$,we get $A = \frac{1}{10}$ and $B = -\frac{1}{15}$.
Therefore,$A - B = \frac{1}{10} - (-\frac{1}{15}) = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$.

(C) Let $I = \int \frac{dx}{3 \sin x - \cos x + 3}$.
Using the substitution $\tan \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.
Substituting these into the integral:
$I = \int \frac{1}{3(\frac{2t}{1+t^2}) - (\frac{1-t^2}{1+t^2}) + 3} \cdot \frac{2 dt}{1+t^2}$
$I = \int \frac{2 dt}{6t - 1 + t^2 + 3(1+t^2)} = \int \frac{2 dt}{4t^2 + 6t + 2} = \int \frac{dt}{2t^2 + 3t + 1}$.
Factor the denominator: $2t^2 + 3t + 1 = (2t+1)(t+1)$.
Using partial fractions: $\frac{1}{(2t+1)(t+1)} = \frac{A}{2t+1} + \frac{B}{t+1}$.
Solving for $A$ and $B$,we get $A = -2$ and $B = 1$.
$I = \int (\frac{1}{t+1} - \frac{2}{2t+1}) dt = \log|t+1| - \log|2t+1| + C = \log \left| \frac{t+1}{2t+1} \right| + C$.
Wait,checking the partial fraction again: $\frac{1}{(2t+1)(t+1)} = \frac{2}{2t+1} - \frac{1}{t+1}$.
$I = \int (\frac{2}{2t+1} - \frac{1}{t+1}) dt = \log|2t+1| - \log|t+1| + C = \log \left| \frac{2t+1}{t+1} \right| + C$.
Substituting $t = \tan \frac{x}{2}$,we get $I = \log \left( \frac{2 \tan \frac{x}{2} + 1}{\tan \frac{x}{2} + 1} \right) + C$.

(A) 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 $t = x^{n}$,then $dt = nx^{n-1} dx$,which implies $x^{n-1} dx = \frac{dt}{n}$.
Substituting these into the integral:
$I = \int \frac{dt/n}{t(t+1)} = \frac{1}{n} \int \frac{dt}{t(t+1)}$.
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:
$I = \frac{1}{n} \log \left( \frac{x^{n}}{x^{n}+1} \right) + C$.

(C) To evaluate the integral $\int \frac{x \, dx}{(x-1)(x-2)}$,we use partial fractions.
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 $x = A(x-2) + B(x-1)$.
Setting $x = 1$,we get $1 = A(1-2) \implies A = -1$.
Setting $x = 2$,we get $2 = B(2-1) \implies B = 2$.
Thus,$\int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$.
Integrating term by term,we get $-\log |x-1| + 2 \log |x-2| + c$.
Using the property of logarithms,$n \log m = \log m^n$,we get $\log |x-2|^2 - \log |x-1| + c = \log \left| \frac{(x-2)^2}{x-1} \right| + c$.

(B) To solve the integral $\int \frac{2x+3}{(x-1)(x^2+1)} dx$,we use partial fractions:
$\frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1}$
$2x+3 = B(x^2+1) + (Cx+D)(x-1)$
Setting $x=1$,we get $2(1)+3 = B(1^2+1) \implies 5 = 2B \implies B = \frac{5}{2}$.
Comparing coefficients of $x^2$: $0 = B+C \implies C = -B = -\frac{5}{2}$.
Comparing constants: $3 = B-D \implies D = B-3 = \frac{5}{2} - 3 = -\frac{1}{2}$.
Thus,the integral becomes $\int \left( \frac{5/2}{x-1} + \frac{-5/2x - 1/2}{x^2+1} \right) dx$.
$= \frac{5}{2} \int \frac{1}{x-1} dx - \frac{5}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
$= \frac{5}{2} \log_e |x-1| - \frac{5}{4} \log_e (x^2+1) - \frac{1}{2} \tan^{-1} x + A$.
$= \log_e |x-1|^{\frac{5}{2}} + \log_e (x^2+1)^{-\frac{5}{4}} - \frac{1}{2} \tan^{-1} x + A$.
$= \log_e \{(x-1)^{\frac{5}{2}}(x^2+1)^{-\frac{5}{4}}\} - \frac{1}{2} \tan^{-1} x + A$.
Comparing this with the given expression,we find $a = -\frac{5}{4}$.

(B) Let $I = \int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx$.
Using partial fractions,let $x^2 = t$. 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 $t=1$,$5 = A(-3) \implies A = -\frac{5}{3}$.
For $t=4$,$11 = B(3) \implies B = \frac{11}{3}$.
So,$\frac{2x^2+3}{(x^2-1)(x^2-4)} = \frac{-5/3}{x^2-1} + \frac{11/3}{x^2-4}$.
Integrating,$I = -\frac{5}{3} \int \frac{dx}{x^2-1} + \frac{11}{3} \int \frac{dx}{x^2-4}$.
$I = -\frac{5}{3} \cdot \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + \frac{11}{3} \cdot \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + c$.
$I = \frac{5}{6} \log \left| \frac{x+1}{x-1} \right| + \frac{11}{12} \log \left| \frac{x-2}{x+2} \right| + c$.
Comparing with the given form,$a = \frac{11}{12}$ and $b = \frac{5}{6} = \frac{10}{12}$.
Thus,$a+b = \frac{11}{12} + \frac{10}{12} = \frac{21}{12}$.

(A) Let $I = \int \frac{dx}{x^4+5x^2+4}$.
Factor the denominator: $x^4+5x^2+4 = (x^2+1)(x^2+4)$.
Using partial fractions: $\frac{1}{(x^2+1)(x^2+4)} = \frac{A'}{x^2+1} + \frac{B'}{x^2+4}$.
$1 = A'(x^2+4) + B'(x^2+1)$.
Setting $x^2 = -1$,we get $1 = A'(3) \implies A' = \frac{1}{3}$.
Setting $x^2 = -4$,we get $1 = B'(-3) \implies B' = -\frac{1}{3}$.
So,$I = \int \left( \frac{1/3}{x^2+1} - \frac{1/3}{x^2+4} \right) dx$.
$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.
$I = \frac{1}{3} \tan^{-1} x - \frac{1}{3} \cdot \frac{1}{2} \tan^{-1} \frac{x}{2} + c$.
$I = \frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \frac{x}{2} + c$.
Comparing with $A \tan^{-1} x + B \tan^{-1} \frac{x}{2} + c$,we get $A = \frac{1}{3}$ and $B = -\frac{1}{6}$.

(A) We have the integral $I = \int \frac{x^4+1}{x(x^2+1)^2} dx$.
Multiply the numerator and denominator by $x$:
$I = \int \frac{x(x^4+1)}{x^2(x^2+1)^2} dx$.
Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{dt}{2}$.
Substituting this into the integral:
$I = \frac{1}{2} \int \frac{t^2+1}{t(t+1)^2} dt$.
Using partial fractions: $\frac{t^2+1}{t(t+1)^2} = \frac{A_1}{t} + \frac{B_1}{t+1} + \frac{C_1}{(t+1)^2}$.
$t^2+1 = A_1(t+1)^2 + B_1 t(t+1) + C_1 t$.
For $t=0$,$1 = A_1(1)^2 \implies A_1 = 1$.
For $t=-1$,$2 = C_1(-1) \implies C_1 = -2$.
Comparing coefficients of $t^2$: $1 = A_1 + B_1 \implies 1 = 1 + B_1 \implies B_1 = 0$.
So,$I = \frac{1}{2} \int \left( \frac{1}{t} - \frac{2}{(t+1)^2} \right) dt = \frac{1}{2} \left( \log |t| + \frac{2}{t+1} \right) + c$.
$I = \frac{1}{2} \log |x^2| + \frac{1}{x^2+1} + c = \log |x| + \frac{1}{x^2+1} + c$.
Comparing with $A \log |x| + \frac{B}{1+x^2} + c$,we get $A = 1$ and $B = 1$.
Thus,$A-B = 1-1 = 0$.

(A) Let $I = \int \frac{2x^2-1}{(x^2+4)(x^2-3)} dx$.
Using partial fractions,let $t = x^2$. Then $\frac{2t-1}{(t+4)(t-3)} = \frac{A}{t+4} + \frac{B}{t-3}$.
$2t-1 = A(t-3) + B(t+4)$.
For $t = 3$,$5 = 7B \Rightarrow B = \frac{5}{7}$.
For $t = -4$,$-9 = -7A \Rightarrow A = \frac{9}{7}$.
Thus,$I = \int \left( \frac{9/7}{x^2+4} + \frac{5/7}{x^2-3} \right) dx$.
$I = \frac{9}{7} \int \frac{1}{x^2+2^2} dx + \frac{5}{7} \int \frac{1}{x^2-(\sqrt{3})^2} dx$.
Using standard integrals $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$ and $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}|$:
$I = \frac{9}{7} \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + \frac{5}{7} \cdot \frac{1}{2\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$.
$I = \frac{9}{14} \tan^{-1}(\frac{x}{2}) + \frac{5}{14\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$.

(D) Let $I = \int \frac{x}{(x-1)^2(x+2)} \, dx$.
Using partial fractions,we write: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.
Multiplying by $(x-1)^2(x+2)$,we get: $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$.
For $x = 1$,$1 = B(1+2) \implies 3B = 1 \implies B = \frac{1}{3}$.
For $x = -2$,$-2 = C(-2-1)^2 \implies -2 = 9C \implies C = -\frac{2}{9}$.
Comparing the coefficients of $x^2$ on both sides: $0 = A + C \implies A = -C = \frac{2}{9}$.
Thus,$I = \int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right) \, dx$.
Integrating term by term: $I = \frac{2}{9} \log |x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log |x+2| + c$.

(B) Let $ I = \int \frac{x^2+1}{x(x^2-1)} dx $.
Divide the numerator and denominator by $ x^2 $:
$ I = \int \frac{1 + \frac{1}{x^2}}{x(1 - \frac{1}{x^2})} dx $.
Let $ t = x - \frac{1}{x} $. Then $ dt = (1 + \frac{1}{x^2}) dx $.
However,the given integral is $ \int \frac{x^2+1}{x^3-x} dx $.
Alternatively,using partial fractions:
$ \frac{x^2+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} $.
$ x^2+1 = A(x^2-1) + Bx(x+1) + Cx(x-1) $.
For $ x=0, 1 = -A \Rightarrow A = -1 $.
For $ x=1, 2 = 2B \Rightarrow B = 1 $.
For $ x=-1, 2 = 2C \Rightarrow C = 1 $.
$ I = \int (-\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx $.
$ I = -\log |x| + \log |x-1| + \log |x+1| + c $.
$ I = \log \left| \frac{(x-1)(x+1)}{x} \right| + c = \log \left| \frac{x^2-1}{x} \right| + c $.

(B) To solve $\int \frac{1}{(x+2)(x^2+1)} \, dx$,we use partial fractions.
Let $\frac{1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$.
Multiplying by $(x+2)(x^2+1)$,we get $1 = A(x^2+1) + (Bx+C)(x+2)$.
Setting $x = -2$,we get $1 = A(4+1) \implies A = \frac{1}{5}$.
Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = -\frac{1}{5}$.
Comparing constants: $1 = A + 2C \implies 1 = \frac{1}{5} + 2C \implies 2C = \frac{4}{5} \implies C = \frac{2}{5}$.
Thus,the integral becomes $\int \left( \frac{1/5}{x+2} + \frac{-1/5x + 2/5}{x^2+1} \right) \, dx$.
$= \frac{1}{5} \int \frac{1}{x+2} \, dx - \frac{1}{10} \int \frac{2x}{x^2+1} \, dx + \frac{2}{5} \int \frac{1}{x^2+1} \, dx$.
$= \frac{1}{5} \log |x+2| - \frac{1}{10} \log |x^2+1| + \frac{2}{5} \tan^{-1} x + c$.
$= \frac{1}{5} \log \left| \frac{x+2}{\sqrt{x^2+1}} \right| + \frac{2}{5} \tan^{-1} x + c$.

(A) To evaluate the integral $\int \frac{3x-2}{(x+1)(x-2)^2} dx$,we use partial fractions.
Let $\frac{3x-2}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$.
Multiplying by $(x+1)(x-2)^2$,we get $3x-2 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$.
Setting $x = -1$: $3(-1)-2 = A(-1-2)^2 \implies -5 = 9A \implies A = -\frac{5}{9}$.
Setting $x = 2$: $3(2)-2 = C(2+1) \implies 4 = 3C \implies C = \frac{4}{3}$.
Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = \frac{5}{9}$.
Thus,$\int \left( \frac{-5/9}{x+1} + \frac{5/9}{x-2} + \frac{4/3}{(x-2)^2} \right) dx = -\frac{5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$.

(C) Let $I = \int \frac{2 x^2-1}{x^4-x^2-20} d x$.
Substitute $x^2 = t$,then the integrand becomes $\frac{2 t-1}{t^2-t-20} = \frac{2 t-1}{(t-5)(t+4)}$.
Using partial fractions,let $\frac{2 t-1}{(t-5)(t+4)} = \frac{A}{t-5} + \frac{B}{t+4}$.
Then $2t-1 = A(t+4) + B(t-5)$.
For $t=5$,$9 = 9A \implies A=1$.
For $t=-4$,$-9 = -9B \implies B=1$.
Thus,$I = \int \left( \frac{1}{x^2-5} + \frac{1}{x^2+4} \right) d x$.
Using the standard integrals $\int \frac{1}{x^2-a^2} d x = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + c$ and $\int \frac{1}{x^2+a^2} d x = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + c$,we get:
$I = \frac{1}{2\sqrt{5}} \log \left| \frac{x-\sqrt{5}}{x+\sqrt{5}} \right| + \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + c$.

(A) Let $I = \int \frac{x^{2}}{(x+1)(x+2)^{2}} dx$.
Using partial fractions,we write:
$\frac{x^{2}}{(x+1)(x+2)^{2}} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^{2}}$.
Multiplying both sides by $(x+1)(x+2)^{2}$,we get:
$x^{2} = A(x+2)^{2} + B(x+1)(x+2) + C(x+1)$.
Setting $x = -2$,we get $(-2)^{2} = C(-2+1) \Rightarrow 4 = -C \Rightarrow C = -4$.
Setting $x = -1$,we get $(-1)^{2} = A(-1+2)^{2} \Rightarrow 1 = A(1)^{2} \Rightarrow A = 1$.
Comparing the coefficients of $x^{2}$ on both sides,we get $1 = A + B$.
Since $A = 1$,we have $1 = 1 + B \Rightarrow B = 0$.
Thus,the integral becomes:
$I = \int \left( \frac{1}{x+1} - \frac{4}{(x+2)^{2}} \right) dx$.
$I = \int \frac{1}{x+1} dx - 4 \int (x+2)^{-2} dx$.
$I = \log |x+1| - 4 \left( \frac{(x+2)^{-1}}{-1} \right) + c$.
$I = \log |x+1| + \frac{4}{x+2} + c$.

(A) Given integral: $I = \int \frac{x^{2}+1}{(x-3)(x-2)} d x$
Since the degree of the numerator is equal to the degree of the denominator,we perform division:
$\frac{x^{2}+1}{x^{2}-5x+6} = 1 + \frac{5x-5}{x^{2}-5x+6}$
So,$I = \int (1 + \frac{5x-5}{(x-3)(x-2)}) d x = \int 1 d x + 5 \int \frac{x-1}{(x-3)(x-2)} d x$
Using partial fractions for $\frac{x-1}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$:
$x-1 = A(x-2) + B(x-3)$
For $x=3$: $3-1 = A(3-2) \Rightarrow A=2$
For $x=2$: $2-1 = B(2-3) \Rightarrow B=-1$
Thus,$I = x + 5 \int (\frac{2}{x-3} - \frac{1}{x-2}) d x = x + 10 \log |x-3| - 5 \log |x-2| + c$
Comparing with $Px + Q \log |x-3| + R \log |x-2| + c$,we get $P=1, Q=10, R=-5$.

(A) Let $I = \int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x$.
Using partial fractions,let $x^2 = t$. 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 $t=1$,$5 = 5A \implies A=1$.
For $t=-4$,$-5 = -5B \implies B=1$.
Thus,$\frac{2x^2+3}{(x^2-1)(x^2+4)} = \frac{1}{x^2-1} + \frac{1}{x^2+4}$.
Integrating both sides,$I = \int \frac{1}{x^2-1} dx + \int \frac{1}{x^2+2^2} dx$.
$I = \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C$.
Comparing this with the given expression $a \log \left| \frac{x-1}{x+1} \right| + b \tan^{-1} \left( \frac{x}{2} \right) + C$,we get $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

(A) Let $I = \int \frac{x+1}{x^{2}+5x+6} dx$.
Factorizing the denominator: $x^{2}+5x+6 = (x+2)(x+3)$.
Using partial fractions: $\frac{x+1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$.
$x+1 = A(x+3) + B(x+2)$.
Setting $x = -2$: $-2+1 = A(-2+3) \implies A = -1$.
Setting $x = -3$: $-3+1 = B(-3+2) \implies -2 = B(-1) \implies B = 2$.
Thus,$I = \int \left( \frac{-1}{x+2} + \frac{2}{x+3} \right) dx$.
$I = -\int \frac{1}{x+2} dx + 2 \int \frac{1}{x+3} dx$.
$I = -\log |x+2| + 2 \log |x+3| + C$.

(D) Let $I = \int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$I = \int \frac{\cos \theta}{5+7 \sin \theta-2(1-\sin^2 \theta)} d \theta = \int \frac{\cos \theta}{2 \sin^2 \theta+7 \sin \theta+3} d \theta$.
Factoring the denominator:
$2 \sin^2 \theta+7 \sin \theta+3 = (2 \sin \theta+1)(\sin \theta+3)$.
Let $\sin \theta = t$,then $\cos \theta d \theta = dt$.
$I = \int \frac{dt}{(2t+1)(t+3)}$.
Using partial fractions:
$\frac{1}{(2t+1)(t+3)} = \frac{A'}{2t+1} + \frac{B'}{t+3} \Rightarrow 1 = A'(t+3) + B'(2t+1)$.
For $t = -3$,$1 = B'(-5) \Rightarrow B' = -\frac{1}{5}$.
For $t = -\frac{1}{2}$,$1 = A'(\frac{5}{2}) \Rightarrow A' = \frac{2}{5}$.
$I = \int (\frac{2/5}{2t+1} - \frac{1/5}{t+3}) dt = \frac{1}{5} \ln |2t+1| - \frac{1}{5} \ln |t+3| + c = \frac{1}{5} \ln |\frac{2 \sin \theta+1}{\sin \theta+3}| + c$.
Comparing with $A \ln |f(\theta)| + c$,we have $A = \frac{1}{5}$ and $f(\theta) = \frac{2 \sin \theta+1}{\sin \theta+3}$.
Thus,$\frac{f(\theta)}{A} = \frac{(2 \sin \theta+1)/(\sin \theta+3)}{1/5} = \frac{5(2 \sin \theta+1)}{\sin \theta+3}$.

(B) Let $I = \int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} dx$.
Since $\tan x = \frac{\sin x}{\cos x}$ and $\frac{1}{\cos x} = \sec x$,the integral becomes:
$I = \int \frac{\sec x \tan x}{(\sec x-1)(\sec x-2)} dx$.
Let $u = \sec x$,then $du = \sec x \tan x dx$.
The integral transforms to:
$I = \int \frac{1}{(u-1)(u-2)} du$.
Using partial fractions:
$\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}$.
$1 = A(u-2) + B(u-1)$.
For $u=1$,$A = -1$. For $u=2$,$B = 1$.
So,$I = \int \left( \frac{1}{u-2} - \frac{1}{u-1} \right) du$.
$I = \log |u-2| - \log |u-1| + c = \log \left| \frac{u-2}{u-1} \right| + c$.
Substituting $u = \sec x$,we get:
$I = \log \left| \frac{\sec x - 2}{\sec x - 1} \right| + c$.