Partial fractions Questions in English

(B) Given the partial fraction decomposition: $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$.
Multiplying both sides by $(x + 1)(x - 3)$,we get: $2x + 3 = a(x - 3) + b(x + 1)$.
To find $a$,set $x = -1$: $2(-1) + 3 = a(-1 - 3)$ $\Rightarrow 1 = -4a$ $\Rightarrow a = -\frac{1}{4}$.
To find $b$,set $x = 3$: $2(3) + 3 = b(3 + 1)$ $\Rightarrow 9 = 4b$ $\Rightarrow b = \frac{9}{4}$.
Therefore,$a + b = -\frac{1}{4} + \frac{9}{4} = \frac{8}{4} = 2$.

(D) Given the equation: $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$
Since $x^2 - 3x + 2 = (x - 2)(x - 1)$,we can write:
$\frac{3x + a}{(x - 2)(x - 1)} = \frac{A(x - 1) - 10(x - 2)}{(x - 2)(x - 1)}$
Equating the numerators:
$3x + a = A(x - 1) - 10(x - 2)$
$3x + a = Ax - A - 10x + 20$
$3x + a = (A - 10)x + (20 - A)$
Comparing the coefficients of $x$ and the constant terms:
$3 = A - 10 \implies A = 13$
$a = 20 - A \implies a = 20 - 13 = 7$
Thus,both $a = 7$ and $A = 13$ are correct. Therefore,the correct option is $(d)$.

(C) Given the partial fraction decomposition: $\frac{3x + 4}{{(x + 1)}^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^2}$
Multiplying both sides by the denominator ${(x + 1)}^2(x - 1)$,we get:
$3x + 4 = A{(x + 1)}^2 + B(x + 1)(x - 1) + C(x - 1)$
To find $A$,substitute $x = 1$ into the equation:
$3(1) + 4 = A(1 + 1)^2 + B(1 + 1)(1 - 1) + C(1 - 1)$
$7 = A(2)^2 + 0 + 0$
$7 = 4A$
$A = \frac{7}{4} = 1.75$

(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.
Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:
$3x - 1 = (Ax + B)(x + 2) + C(x^2 - x + 1)$.
For $x = -2$:
$3(-2) - 1 = C((-2)^2 - (-2) + 1)$
$-7 = C(4 + 2 + 1) \implies -7 = 7C \implies C = -1$.
Comparing coefficients of $x^2$:
$0 = A + C \implies A = -C = 1$.
Comparing constant terms:
$-1 = 2B + C \implies -1 = 2B - 1 \implies 2B = 0 \implies B = 0$.
Substituting $A=1, B=0, C=-1$ into the partial fraction form:
$\frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$.

(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \left[ \frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} \right] + y$
We express the partial fraction decomposition of the left side as: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{(x^2 + 1)^2}$
Comparing the given expression with the standard decomposition,we identify $y$ as the term $\frac{Dx + E}{(x^2 + 1)^2}$.
Multiplying by the common denominator $(x - 1)(x^2 + 1)^2$:
$x = \frac{1}{4}(x^2 + 1)^2 - \frac{1}{4}(x + 1)(x - 1)(x^2 + 1) + y(x - 1)(x^2 + 1)^2$
Solving for the coefficients,we find $y = \frac{1 - x}{2(x^2 + 1)^2}$.

(A) Using partial fractions,we write: $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{A}{2 + x} + \frac{B}{1 - x}$.
Solving for $A$ and $B$: $5x + 6 = A(1 - x) + B(2 + x)$.
For $x = 1$: $11 = 3B \implies B = \frac{11}{3}$.
For $x = -2$: $-4 = 3A \implies A = -\frac{4}{3}$.
Thus,the expression is $\frac{-4/3}{2 + x} + \frac{11/3}{1 - x} = \frac{-2/3}{1 + x/2} + \frac{11/3}{1 - x}$.
Expanding using the binomial series $(1 + z)^{-1} = \sum_{n=0}^{\infty} (-1)^n z^n$ and $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$:
$= -\frac{2}{3} \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{2}\right)^n + \frac{11}{3} \sum_{n=0}^{\infty} x^n$.
The coefficient of $x^n$ is $-\frac{2}{3} \cdot \frac{(-1)^n}{2^n} + \frac{11}{3}$.

(B) Given the partial fraction decomposition: $\frac{1}{x(x + 1)...(x + n)} = \sum_{r=0}^{n} \frac{A_r}{x + r}$.
To find $A_r$,multiply both sides by $(x + r)$ and take the limit as $x \to -r$:
$A_r = \lim_{x \to -r} \frac{x + r}{x(x + 1)...(x + n)}$.
This is equivalent to $\frac{1}{\prod_{k=0, k \neq r}^{n} (-r + k)}$.
The product in the denominator is: $(-r)(-r+1)...(-1) \times (1)(2)...(n-r)$.
This simplifies to $(-1)^r \times r! \times (n-r)!$.
Therefore,$A_r = \frac{1}{(-1)^r r! (n-r)!} = \frac{(-1)^r}{r!(n-r)!}$.

(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.
Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,we get:
$x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$.
For $x = 1$: $1 + 1 = A(1 - 2)(1 - 3)$ $\Rightarrow 2 = A(-1)(-2)$ $\Rightarrow 2 = 2A$ $\Rightarrow A = 1$.
For $x = 2$: $2 + 1 = B(2 - 1)(2 - 3)$ $\Rightarrow 3 = B(1)(-1)$ $\Rightarrow 3 = -B$ $\Rightarrow B = -3$.
For $x = 3$: $3 + 1 = C(3 - 1)(3 - 2)$ $\Rightarrow 4 = C(2)(1)$ $\Rightarrow 4 = 2C$ $\Rightarrow C = 2$.
Therefore,the partial fraction decomposition is $\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3}$.

(A) Given $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$.
Multiplying both sides by the denominator $(x - 1)(x + 2)(2x + 3)$,we get:
$ax^2 + bx + c = 3(x + 2)(2x + 3) + 2(x - 1)(2x + 3) - 5(x - 1)(x + 2)$.
Expanding the terms:
$3(2x^2 + 7x + 6) + 2(2x^2 + x - 3) - 5(x^2 + x - 2)$
$= (6x^2 + 21x + 18) + (4x^2 + 2x - 6) - (5x^2 + 5x - 10)$
$= (6 + 4 - 5)x^2 + (21 + 2 - 5)x + (18 - 6 + 10)$
$= 5x^2 + 18x + 22$.
Comparing coefficients,we get $a = 5$,$b = 18$,and $c = 22$.
Thus,$a = 5$ is correct.

(D) Given the partial fraction decomposition: $\frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -\frac{3}{e^x - 1} + \frac{B}{2e^x - 3}$
Multiply both sides by $(e^x - 1)(2e^x - 3)$ to clear the denominators:
$e^x + 2 = -3(2e^x - 3) + B(e^x - 1)$
$e^x + 2 = -6e^x + 9 + Be^x - B$
$e^x + 2 = (B - 6)e^x + (9 - B)$
Comparing the coefficients of $e^x$ and the constant terms:
$B - 6 = 1 \Rightarrow B = 7$
$9 - B = 2 \Rightarrow B = 7$
Thus,$B = 7$.

(B) Given the partial fraction decomposition: $\frac{3x + 4}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{B}{x - 1}$.
Multiplying both sides by $(x - 2)(x - 1)$,we get: $3x + 4 = A(x - 1) - B(x - 2)$.
Expanding the right side: $3x + 4 = Ax - A - Bx + 2B = (A - B)x + (-A + 2B)$.
Comparing the coefficients of $x$ and the constant terms:
$A - B = 3$ (Equation $1$)
$-A + 2B = 4$ (Equation $2$)
Adding Equation $1$ and Equation $2$: $(A - B) + (-A + 2B) = 3 + 4 \Rightarrow B = 7$.
Substituting $B = 7$ into Equation $1$: $A - 7 = 3 \Rightarrow A = 10$.
Therefore,$(A, B) = (10, 7)$.

(D) Given the equation: $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$
Simplify the right side by taking a common denominator:
$\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2(x^2 + b^2) - b^2(x^2 + a^2)}{(x^2 + a^2)(x^2 + b^2)} \right)$
Equating the numerators:
$x^2 = k [a^2 x^2 + a^2 b^2 - b^2 x^2 - a^2 b^2]$
$x^2 = k [a^2 x^2 - b^2 x^2]$
$x^2 = k x^2 (a^2 - b^2)$
Dividing both sides by $x^2$:
$1 = k(a^2 - b^2)$
Therefore,$k = \frac{1}{a^2 - b^2}$.

(C) Given the partial fraction decomposition: $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$.
Multiplying both sides by $(x - 1)(x + 2)^2$,we get: $9 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1)$.
For $x = 1$: $9 = A(1 + 2)^2$ $\Rightarrow 9 = 9A$ $\Rightarrow A = 1$.
For $x = -2$: $9 = C(-2 - 1)$ $\Rightarrow 9 = -3C$ $\Rightarrow C = -3$.
Equating the coefficients of $x^2$ on both sides: $0 = A + B \Rightarrow B = -A = -1$.
Therefore,$A - B - C = 1 - (-1) - (-3) = 1 + 1 + 3 = 5$.

(D) Given the equation: $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$
To solve for $a$ and $b$,multiply both sides by $(3x + 4)^2$:
$ax + b = (3x + 4) - 3$
$ax + b = 3x + 1$
By comparing the coefficients of $x$ and the constant terms on both sides,we get:
$a = 3$
$b = 1$
Thus,both options $(b)$ and $(c)$ are correct.

(A) Let $\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{A}{2x + 3} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}$.
Multiplying both sides by $(2x + 3)(x + 3)^2$,we get:
$x^2 + 13x + 15 = A(x + 3)^2 + B(2x + 3)(x + 3) + C(2x + 3)$.
For $x = -3$,$(-3)^2 + 13(-3) + 15 = C(2(-3) + 3)$ $\Rightarrow 9 - 39 + 15 = -3C$ $\Rightarrow -15 = -3C$ $\Rightarrow C = 5$.
For $x = -\frac{3}{2}$,$(-\frac{3}{2})^2 + 13(-\frac{3}{2}) + 15 = A(-\frac{3}{2} + 3)^2$ $\Rightarrow \frac{9}{4} - \frac{39}{2} + 15 = A(\frac{3}{2})^2$ $\Rightarrow \frac{9 - 78 + 60}{4} = \frac{9}{4}A$ $\Rightarrow -9 = 9A$ $\Rightarrow A = -1$.
Equating the coefficients of $x^2$ on both sides:
$1 = A + 2B$ $\Rightarrow 1 = -1 + 2B$ $\Rightarrow 2B = 2$ $\Rightarrow B = 1$.
Substituting the values of $A, B, C$ back into the partial fraction form:
$\frac{1}{x + 3} - \frac{1}{2x + 3} + \frac{5}{(x + 3)^2}$.

(C) Let $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{(x - 1)^4}$.
Multiplying both sides by $(x - 1)^4$,we get:
$3x^3 - 8x^2 + 10 = A(x - 1)^3 + B(x - 1)^2 + C(x - 1) + D$.
Expanding the terms:
$3x^3 - 8x^2 + 10 = A(x^3 - 3x^2 + 3x - 1) + B(x^2 - 2x + 1) + C(x - 1) + D$.
Comparing coefficients of powers of $x$:
Coefficient of $x^3$: $A = 3$.
Coefficient of $x^2$: $-3A + B = -8$ $\Rightarrow -3(3) + B = -8$ $\Rightarrow B = 1$.
Coefficient of $x$: $3A - 2B + C = 0$ $\Rightarrow 3(3) - 2(1) + C = 0$ $\Rightarrow 9 - 2 + C = 0$ $\Rightarrow C = -7$.
Constant term: $-A + B - C + D = 10$ $\Rightarrow -3 + 1 - (-7) + D = 10$ $\Rightarrow 5 + D = 10$ $\Rightarrow D = 5$.
Thus,the partial fractions are $\frac{3}{x - 1} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4}$.

(B) Given the partial fraction decomposition: $\frac{(x - 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.
Multiplying both sides by $x(x^2 + 1)$,we get: $(x - 1)^2 = A(x^2 + 1) + x(Bx + C)$.
Expanding the left side: $x^2 - 2x + 1 = Ax^2 + A + Bx^2 + Cx$.
Grouping terms by powers of $x$: $x^2 - 2x + 1 = (A + B)x^2 + Cx + A$.
Comparing coefficients on both sides:
$1$) Constant term: $A = 1$.
$2$) Coefficient of $x$: $C = -2$.
$3$) Coefficient of $x^2$: $A + B = 0$. Since $A = 1$,then $1 + B = 0$,which gives $B = -1$.
Wait,let us re-evaluate: $A + B = 1$. Since $A = 1$,then $1 + B = 1$,which gives $B = 0$.
Thus,$A = 1, B = 0, C = -2$.

(D) Given $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.
Multiplying both sides by $(x^3 - 1) = (x - 1)(x^2 + x + 1)$,we get:
$2x = A(x^2 + x + 1) + (Bx + C)(x - 1)$.
For $x = 1$,$2(1) = A(1 + 1 + 1) + 0$ $\Rightarrow 3A = 2$ $\Rightarrow A = \frac{2}{3}$.
Expanding the $RHS$: $2x = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.
$2x = (A + B)x^2 + (A - B + C)x + (A - C)$.
Comparing coefficients of $x^2$: $A + B = 0 \Rightarrow B = -A = -\frac{2}{3}$.
Comparing constant terms: $A - C = 0 \Rightarrow C = A = \frac{2}{3}$.
Thus,$A = \frac{2}{3}$,$B = -\frac{2}{3}$,and $C = \frac{2}{3}$.
Since $A = C = \frac{2}{3}$ and $B = -\frac{2}{3}$,we have $A \neq B$ and $B \neq C$ and $A \neq B \neq C$ is satisfied as $A \neq B$ and $B \neq C$ and $A \neq C$ is false,but the option $A \neq B \neq C$ is interpreted as $A \neq B$ and $B \neq C$ and $A \neq C$. However,since $A=C$,the condition $A \neq B \neq C$ is the best fit among the choices provided.

(B) We express the given fraction as partial fractions:
$\frac{x^2 + 1}{(2x - 1)(x^2 - 1)} = \frac{A}{2x - 1} + \frac{B}{x + 1} + \frac{C}{x - 1}$
Multiplying both sides by $(2x - 1)(x^2 - 1)$,we get:
$x^2 + 1 = A(x^2 - 1) + B(2x - 1)(x - 1) + C(2x - 1)(x + 1)$
For $x = 1$:
$1^2 + 1 = C(2(1) - 1)(1 + 1)$ $\Rightarrow 2 = 2C$ $\Rightarrow C = 1$
For $x = -1$:
$(-1)^2 + 1 = B(2(-1) - 1)(-1 - 1)$ $\Rightarrow 2 = B(-3)(-2)$ $\Rightarrow 2 = 6B$ $\Rightarrow B = \frac{1}{3}$
For $x = \frac{1}{2}$:
$(\frac{1}{2})^2 + 1 = A((\frac{1}{2})^2 - 1)$ $\Rightarrow \frac{5}{4} = A(\frac{1}{4} - 1)$ $\Rightarrow \frac{5}{4} = -\frac{3}{4}A$ $\Rightarrow A = -\frac{5}{3}$
Substituting the values of $A, B,$ and $C$:
$\frac{-5}{3(2x - 1)} + \frac{1}{3(x + 1)} + \frac{1}{x - 1}$

(B) Given the equation: $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$
Multiply both sides by the common denominator $(1 - x + x^2)(2 + x)$:
$ax - 1 = x(2 + x) - 1(1 - x + x^2)$
Expand the right side:
$ax - 1 = 2x + x^2 - 1 + x - x^2$
Simplify the expression:
$ax - 1 = 3x - 1$
Comparing the coefficients of $x$ on both sides,we get:
$a = 3$

(A) Given the partial fraction decomposition: $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$
Multiplying both sides by $x(x^2 + 1)$,we get: $1 = A(x^2 + 1) + (Bx + C)x$
$1 = Ax^2 + A + Bx^2 + Cx$
$1 = (A + B)x^2 + Cx + A$
Comparing the coefficients of $x^2$,$x$,and the constant term on both sides:
$A + B = 0$
$C = 0$
$A = 1$
Substituting $A = 1$ into $A + B = 0$,we get $1 + B = 0$,which implies $B = -1$.
Thus,$(A, B, C) = (1, -1, 0)$.

(D) We start with the expression $\frac{2x}{x^4 + x^2 + 1}$.
First,factor the denominator: $x^4 + x^2 + 1 = (x^4 + 2x^2 + 1) - x^2 = (x^2 + 1)^2 - x^2$.
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$,we get $(x^2 - x + 1)(x^2 + x + 1)$.
Now,express the fraction as: $\frac{2x}{(x^2 - x + 1)(x^2 + x + 1)}$.
Using partial fractions,we set $\frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{A}{x^2 - x + 1} + \frac{B}{x^2 + x + 1}$.
By solving,we find the difference: $\frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1} = \frac{(x^2 + x + 1) - (x^2 - x + 1)}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{2x}{x^4 + x^2 + 1}$.
Thus,the correct option is $D$.

(B) Given the partial fraction decomposition: $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$
Multiplying both sides by $(x^2 + 1)^2$,we get: $3x^2 + 5 = a(x^2 + 1) + b$
Expanding the right side: $3x^2 + 5 = ax^2 + a + b$
Comparing the coefficients of $x^2$ and the constant terms on both sides:
$a = 3$
$a + b = 5$
Substituting $a = 3$ into the second equation: $3 + b = 5 \Rightarrow b = 2$
Therefore,$(a, b) = (3, 2)$.

(D) Given the equation: $\frac{(x - a)(x - d)}{(x - c)(x - d)} = \frac{A}{x - c} - \frac{B}{x - d} + C$
Multiply both sides by $(x - c)(x - d)$:
$(x - a)(x - b) = A(x - d) - B(x - c) + C(x - c)(x - d)$
Expanding the right side:
$x^2 - (a + b)x + ab = Ax - Ad - Bx + Bc + C(x^2 - (c + d)x + cd)$
Equating the coefficients of $x^2$ on both sides:
$1 = C$
Therefore,$C = 1$.

(D) Given expression: $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.
Since the degree of the numerator is equal to the degree of the denominator,we perform long division:
$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{3x - 7}{x^2 - 3x + 2}$.
Now,factor the denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$.
We express the remainder as partial fractions: $\frac{3x - 7}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.
$3x - 7 = A(x - 2) + B(x - 1)$.
Setting $x = 1$: $3(1) - 7 = A(1 - 2)$ $\Rightarrow -4 = -A$ $\Rightarrow A = 4$.
Setting $x = 2$: $3(2) - 7 = B(2 - 1) \Rightarrow -1 = B$.
Thus,$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{4}{x - 1} - \frac{1}{x - 2}$.

(A) Divide the numerator by the denominator: $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{6x^2 + 5x + 1} = x^2 + \frac{5x + 2}{6x^2 + 5x + 1}$.
Factor the denominator: $6x^2 + 5x + 1 = (2x + 1)(3x + 1)$.
Now,express $\frac{5x + 2}{(2x + 1)(3x + 1)}$ as partial fractions: $\frac{5x + 2}{(2x + 1)(3x + 1)} = \frac{A}{2x + 1} + \frac{B}{3x + 1}$.
$5x + 2 = A(3x + 1) + B(2x + 1)$.
For $x = -\frac{1}{2}$,$5(-\frac{1}{2}) + 2 = A(3(-\frac{1}{2}) + 1) \implies -\frac{1}{2} = A(-\frac{1}{2}) \implies A = 1$.
For $x = -\frac{1}{3}$,$5(-\frac{1}{3}) + 2 = B(2(-\frac{1}{3}) + 1) \implies \frac{1}{3} = B(\frac{1}{3}) \implies B = 1$.
Thus,the expression is $x^2 + \frac{1}{2x + 1} + \frac{1}{3x + 1}$.

(D) Let $u = \sin x$. The expression is $\frac{u^2 + 1}{2u^2 - 5u + 3} = \frac{u^2 + 1}{(2u - 3)(u - 1)}$.
Using partial fractions,we write $\frac{u^2 + 1}{(2u - 3)(u - 1)} = C + \frac{A}{2u - 3} + \frac{B}{u - 1}$.
Since the degree of the numerator equals the degree of the denominator,$C$ is the ratio of the leading coefficients: $C = \frac{1}{2}$.
Now,$\frac{u^2 + 1}{(2u - 3)(u - 1)} = \frac{1}{2} + \frac{A}{2u - 3} + \frac{B}{u - 1}$.
$\frac{u^2 + 1 - \frac{1}{2}(2u^2 - 5u + 3)}{(2u - 3)(u - 1)} = \frac{\frac{5}{2}u - \frac{1}{2}}{(2u - 3)(u - 1)} = \frac{A}{2u - 3} + \frac{B}{u - 1}$.
Using the cover-up method:
For $B$: Multiply by $(u - 1)$ and set $u = 1$: $B = \frac{\frac{5}{2}(1) - \frac{1}{2}}{2(1) - 3} = \frac{2}{-1} = -2$.
For $A$: Multiply by $(2u - 3)$ and set $u = \frac{3}{2}$: $A = \frac{\frac{5}{2}(\frac{3}{2}) - \frac{1}{2}}{\frac{3}{2} - 1} = \frac{\frac{15}{4} - \frac{2}{4}}{\frac{1}{2}} = \frac{13}{4} \times 2 = \frac{13}{2}$.
Thus,$A = \frac{13}{2}$,$B = -2$,$C = \frac{1}{2}$.
$A + B + C = \frac{13}{2} - 2 + \frac{1}{2} = 7 - 2 = 5$.
Both $(A)$ and $(B)$ are correct.

(B) Using partial fractions,$\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$.
Solving for $A$ and $B$,we get $A = 2$ and $B = 1$,so $\frac{3x}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{1}{x + 1} = -\frac{1}{1 - x/2} + \frac{1}{1 + x}$.
Expanding both terms using the binomial series $(1 - z)^{-1} = 1 + z + z^2 + z^3 + z^4 + \dots$ and $(1 + z)^{-1} = 1 - z + z^2 - z^3 + z^4 - \dots$:
$-\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \dots\right) + \left(1 - x + x^2 - x^3 + x^4 - \dots\right)$.
The coefficient of $x^4$ is $-\frac{1}{16} + 1 = \frac{15}{16}$.

(D) Let $\frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$.
By partial fraction decomposition,$x^2 + 1 = (Ax + B)(x - 2) + C(x^2 + 4)$.
Comparing coefficients,we get $A + C = 1$,$-2A + B = 0$,and $-2B + 4C = 1$.
Solving these,we find $A = 3/8$,$B = 3/4$,and $C = 5/8$.
Thus,the expression is $\frac{3x/8 + 3/4}{x^2 + 4} + \frac{5/8}{x - 2}$.
Rewriting for expansion: $\frac{1}{4}(\frac{3x}{8} + \frac{3}{4})(1 + \frac{x^2}{4})^{-1} - \frac{5}{16}(1 - \frac{x}{2})^{-1}$.
Using the binomial expansion $(1+y)^{-1} = 1 - y + y^2 - y^3 + \dots$ and $(1-y)^{-1} = 1 + y + y^2 + y^3 + \dots$:
Term from first part: $\frac{1}{4}(\frac{3x}{8} + \frac{3}{4})(1 - \frac{x^2}{4} + \frac{x^4}{16} - \dots)$. The $x^5$ term comes from $\frac{3x}{8} \times (\frac{x^4}{16}) = \frac{3x^5}{128}$. Multiplying by $1/4$ gives $\frac{3x^5}{512}$.
Term from second part: $-\frac{5}{16}(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \frac{x^5}{32} + \dots)$. The $x^5$ term is $-\frac{5}{16} \times \frac{x^5}{32} = -\frac{5x^5}{512}$.
Summing the coefficients: $\frac{3}{512} - \frac{5}{512} = -\frac{2}{512} = -\frac{1}{256}$.

(C) Let $x - 1 = y$,so $x = y + 1$.
Then,$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = -\frac{1 + 2y + y^2}{y^3(1 - y)}$.
Dividing $(1 + 2y + y^2)$ by $(y - 1)$ gives $(y^2 + 2y + 1) = (y - 1)(y + 3) + 4$.
Thus,$\frac{y^2 + 2y + 1}{y^3(y - 1)} = \frac{(y - 1)(y + 3) + 4}{y^3(y - 1)} = \frac{y + 3}{y^3} + \frac{4}{y^3(y - 1)}$.
Using partial fractions for $\frac{4}{y^3(y - 1)} = \frac{A}{y} + \frac{B}{y^2} + \frac{C}{y^3} + \frac{D}{y - 1}$,we find $D = 4$,$C = -4$,$B = -4$,$A = -4$.
Substituting back,the expression becomes $\frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{(x - 1)} + \frac{4}{(x - 2)}$.

(A) Given the expression $\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6}$.
First,perform polynomial long division of the numerator by the denominator.
The denominator is $x^2 - 5x + 6 = (x - 2)(x - 3)$.
Dividing $x^3 - 6x^2 + 10x - 2$ by $x^2 - 5x + 6$:
$x^3 - 6x^2 + 10x - 2 = x(x^2 - 5x + 6) - x^2 + 4x - 2$
$= x(x^2 - 5x + 6) - 1(x^2 - 5x + 6) - x + 4$
$= (x - 1)(x^2 - 5x + 6) - x + 4$.
Thus,$\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = (x - 1) + \frac{-x + 4}{(x - 2)(x - 3)}$.
Using partial fractions for $\frac{-x + 4}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$:
$-x + 4 = A(x - 3) + B(x - 2)$.
For $x = 2$: $-2 + 4 = A(2 - 3) \implies 2 = -A \implies A = -2$.
For $x = 3$: $-3 + 4 = B(3 - 2) \implies 1 = B \implies B = 1$.
So,the expression is $(x - 1) - \frac{2}{x - 2} + \frac{1}{x - 3}$.
Comparing this with $f(x) + \frac{A}{x - 2} + \frac{B}{x - 3}$,we get $f(x) = x - 1$.

(A) Let $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3} = \frac{A_1 x + B_1}{x^2 + 1} + \frac{A_2 x + B_2}{(x^2 + 1)^2} + \frac{A_3 x + B_3}{(x^2 + 1)^3}$.
Then,$x^4 + 24x^2 + 28 = (A_1 x + B_1)(x^2 + 1)^2 + (A_2 x + B_2)(x^2 + 1) + (A_3 x + B_3)$.
Let $x^2 = y$. Then $\frac{y^2 + 24y + 28}{(y + 1)^3} = \frac{A}{y + 1} + \frac{B}{(y + 1)^2} + \frac{C}{(y + 1)^3}$.
Let $y + 1 = t$,so $y = t - 1$. Then $y^2 + 24y + 28 = (t - 1)^2 + 24(t - 1) + 28 = t^2 - 2t + 1 + 24t - 24 + 28 = t^2 + 22t + 5$.
Thus,$\frac{t^2 + 22t + 5}{t^3} = \frac{1}{t} + \frac{22}{t^2} + \frac{5}{t^3}$.
Substituting $t = x^2 + 1$,we get $\frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$.

(A) We use partial fractions to decompose the expression:
$\frac{1}{(1 - x)(3 - x)} = \frac{A}{1 - x} + \frac{B}{3 - x}$
Solving for $A$ and $B$,we get $1 = A(3 - x) + B(1 - x)$.
For $x = 1$,$1 = 2A \implies A = 1/2$.
For $x = 3$,$1 = -2B \implies B = -1/2$.
Thus,$\frac{1}{(1 - x)(3 - x)} = \frac{1}{2} \left( \frac{1}{1 - x} - \frac{1}{3 - x} \right) = \frac{1}{2} \left[ (1 - x)^{-1} - \frac{1}{3} (1 - x/3)^{-1} \right]$.
Using the binomial expansion $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$:
$\frac{1}{2} \left[ \sum_{n=0}^{\infty} x^n - \frac{1}{3} \sum_{n=0}^{\infty} (x/3)^n \right] = \frac{1}{2} \sum_{n=0}^{\infty} \left( 1 - \frac{1}{3^{n+1}} \right) x^n$.
The coefficient of $x^n$ is $\frac{1}{2} \left( 1 - \frac{1}{3^{n+1}} \right) = \frac{3^{n+1} - 1}{2 \cdot 3^{n+1}}$.

(D) The given function is $f(x) = \frac{1}{(1 - ax)(1 - bx)}$.
Using partial fractions,we can write:
$\frac{1}{(1 - ax)(1 - bx)} = \frac{A}{1 - ax} + \frac{B}{1 - bx}$.
Solving for $A$ and $B$,we get $1 = A(1 - bx) + B(1 - ax)$.
For $x = 1/a$,$1 = A(1 - b/a) \implies A = \frac{a}{a - b}$.
For $x = 1/b$,$1 = B(1 - a/b) \implies B = \frac{b}{b - a}$.
Thus,$f(x) = \frac{a}{a - b}(1 - ax)^{-1} + \frac{b}{b - a}(1 - bx)^{-1}$.
Expanding using the binomial series $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$:
$f(x) = \frac{a}{a - b} \sum_{n=0}^{\infty} (ax)^n + \frac{b}{b - a} \sum_{n=0}^{\infty} (bx)^n$.
The coefficient of $x^n$ is $a_n = \frac{a}{a - b} a^n + \frac{b}{b - a} b^n$.
$a_n = \frac{a^{n+1}}{a - b} + \frac{b^{n+1}}{b - a} = \frac{b^{n+1} - a^{n+1}}{b - a}$.

(D) Given the equation: $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$.
Multiply both sides by $(2x - 1)(x + 2)(x - 3)$:
$x^3 = p(2x - 1)(x + 2)(x - 3) + q(x + 2)(x - 3) + r(2x - 1)(x - 3) + s(2x - 1)(x + 2)$.
Expanding the term with $p$: $p(2x - 1)(x^2 - x - 6) = p(2x^3 - 2x^2 - 12x - x^2 + x + 6) = p(2x^3 - 3x^2 - 11x + 6)$.
Equating the coefficient of $x^3$ on both sides: $1 = 2p \Rightarrow p = \frac{1}{2}$.
Now,to find the constant term,set $x = 0$ in the equation:
$0 = p(-1)(2)(-3) + q(2)(-3) + r(-1)(-3) + s(-1)(2)$.
$0 = 6p - 6q + 3r - 2s$.
Substituting $p = \frac{1}{2}$:
$0 = 6(\frac{1}{2}) - 6q + 3r - 2s$.
$0 = 3 - 6q + 3r - 2s$.
$6q - 3r + 2s = 3$.

(B) Given the partial fraction decomposition: $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$.
Multiplying both sides by $(x + 1)(x - 3)$,we get: $2x + 3 = a(x - 3) + b(x + 1)$.
Setting $x = -1$: $2(-1) + 3 = a(-1 - 3)$ $\Rightarrow 1 = -4a$ $\Rightarrow a = -\frac{1}{4}$.
Setting $x = 3$: $2(3) + 3 = b(3 + 1)$ $\Rightarrow 9 = 4b$ $\Rightarrow b = \frac{9}{4}$.
Therefore,$a + b = -\frac{1}{4} + \frac{9}{4} = \frac{8}{4} = 2$.

(D) Given the equation: $\frac{3x + a}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{10}{x - 1}$.
Multiplying both sides by $(x - 2)(x - 1)$,we get:
$3x + a = A(x - 1) - 10(x - 2)$.
Expanding the right side:
$3x + a = Ax - A - 10x + 20$.
$3x + a = (A - 10)x + (20 - A)$.
Comparing the coefficients of $x$ and the constant terms on both sides:
$3 = A - 10 \implies A = 13$.
$a = 20 - A \implies a = 20 - 13 = 7$.
Thus,both $a = 7$ and $A = 13$ are correct.

(C) Given the partial fraction decomposition: $\frac{3x + 4}{(x + 1)^2(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^2}$
Multiplying both sides by $(x + 1)^2(x - 1)$,we get:
$3x + 4 = A(x + 1)^2 + B(x + 1)(x - 1) + C(x - 1)$
To find $A$,substitute $x = 1$ into the equation:
$3(1) + 4 = A(1 + 1)^2 + B(1 + 1)(1 - 1) + C(1 - 1)$
$7 = A(2)^2 + 0 + 0$
$7 = 4A$
$A = \frac{7}{4}$

(C) Let $(x - 1) = y$,then $x = y + 1$.
Substituting this into the expression:
$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)}$.
Performing division of $(y^2 + 2y + 1)$ by $(y - 1)$:
$\frac{y^2 + 2y + 1}{y - 1} = (y + 3) + \frac{4}{y - 1}$.
Thus,$\frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1}{y^3} \left( \frac{y^2 + 2y + 1}{y - 1} \right) = \frac{1}{y^3} \left( y + 3 + \frac{4}{y - 1} \right) = \frac{1}{y^2} + \frac{3}{y^3} + \frac{4}{y^3(y - 1)}$.
Alternatively,using partial fraction decomposition for $\frac{x^2}{(x - 1)^3(x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2}$.
Solving for coefficients gives $A = -4, B = -3, C = -1, D = 4$.
Therefore,the expression is $\frac{-4}{x - 1} - \frac{3}{(x - 1)^2} - \frac{1}{(x - 1)^3} + \frac{4}{x - 2}$.

(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.
Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:
$3x - 1 = (Ax + B)(x + 2) + C(x^2 - x + 1)$.
To find $C$,put $x = -2$:
$3(-2) - 1 = C((-2)^2 - (-2) + 1) \implies -7 = C(4 + 2 + 1) \implies -7 = 7C \implies C = -1$.
Equating the coefficients of $x^2$:
$0 = A + C \implies A = -C = 1$.
Equating the constant terms:
$-1 = 2B + C \implies -1 = 2B - 1 \implies 2B = 0 \implies B = 0$.
Substituting the values of $A, B, C$:
$\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$.

(A) Using partial fractions,we write: $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{A}{2 + x} + \frac{B}{1 - x}$.
Solving for $A$ and $B$: $5x + 6 = A(1 - x) + B(2 + x)$.
For $x = 1$: $11 = 3B \implies B = \frac{11}{3}$.
For $x = -2$: $-4 = 3A \implies A = -\frac{4}{3}$.
Thus,$\frac{5x + 6}{(2 + x)(1 - x)} = \frac{-4/3}{2 + x} + \frac{11/3}{1 - x} = \frac{-2/3}{1 + x/2} + \frac{11/3}{1 - x}$.
Expanding as a power series: $\frac{-2}{3}(1 + x/2)^{-1} + \frac{11}{3}(1 - x)^{-1}$.
$= \frac{-2}{3} \sum_{n=0}^{\infty} (-\frac{x}{2})^n + \frac{11}{3} \sum_{n=0}^{\infty} x^n$.
$= \sum_{n=0}^{\infty} [\frac{-2}{3} \frac{(-1)^n}{2^n} + \frac{11}{3}] x^n$.
The coefficient of $x^n$ is $\frac{-2}{3} \frac{(-1)^n}{2^n} + \frac{11}{3}$.

(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.
Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,we get:
$x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$.
For $x = 1$: $1 + 1 = A(1 - 2)(1 - 3) \implies 2 = A(-1)(-2) \implies 2 = 2A \implies A = 1$.
For $x = 2$: $2 + 1 = B(2 - 1)(2 - 3) \implies 3 = B(1)(-1) \implies 3 = -B \implies B = -3$.
For $x = 3$: $3 + 1 = C(3 - 1)(3 - 2) \implies 4 = C(2)(1) \implies 4 = 2C \implies C = 2$.
Thus,the partial fraction is $\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3}$.

(A) Let $\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{A}{2x + 3} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}$.
Multiplying both sides by $(2x + 3)(x + 3)^2$,we get:
$x^2 + 13x + 15 = A(x + 3)^2 + B(2x + 3)(x + 3) + C(2x + 3)$.
Setting $x = -3$: $(-3)^2 + 13(-3) + 15 = C(2(-3) + 3) \implies 9 - 39 + 15 = -3C \implies -15 = -3C \implies C = 5$.
Setting $x = -\frac{3}{2}$: $(-\frac{3}{2})^2 + 13(-\frac{3}{2}) + 15 = A(-\frac{3}{2} + 3)^2 \implies \frac{9}{4} - \frac{39}{2} + 15 = A(\frac{3}{2})^2 \implies \frac{9 - 78 + 60}{4} = A(\frac{9}{4}) \implies -9 = 9A \implies A = -1$.
Comparing coefficients of $x^2$: $1 = A + 2B \implies 1 = -1 + 2B \implies 2 = 2B \implies B = 1$.
Thus,the partial fractions are $\frac{1}{x + 3} - \frac{1}{2x + 3} + \frac{5}{(x + 3)^2}$.

(A) Let $u = x - 1$,so $x = u + 1$.
Substituting this into the numerator: $3(u + 1)^3 - 8(u + 1)^2 + 10$.
$= 3(u^3 + 3u^2 + 3u + 1) - 8(u^2 + 2u + 1) + 10$.
$= 3u^3 + 9u^2 + 9u + 3 - 8u^2 - 16u - 8 + 10$.
$= 3u^3 + u^2 - 7u + 5$.
Now,divide by $u^4$: $\frac{3u^3 + u^2 - 7u + 5}{u^4} = \frac{3}{u} + \frac{1}{u^2} - \frac{7}{u^3} + \frac{5}{u^4}$.
Substituting $u = x - 1$ back: $\frac{3}{(x - 1)} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4}$.

(A) Let $u = x^2 + 1$,then $x^2 = u - 1$.
Substituting this into the numerator: $x^4 + 24x^2 + 28 = (u - 1)^2 + 24(u - 1) + 28$.
$= (u^2 - 2u + 1) + 24u - 24 + 28 = u^2 + 22u + 5$.
Now,the expression becomes $\frac{u^2 + 22u + 5}{u^3}$.
$= \frac{u^2}{u^3} + \frac{22u}{u^3} + \frac{5}{u^3} = \frac{1}{u} + \frac{22}{u^2} + \frac{5}{u^3}$.
Substituting $u = x^2 + 1$ back,we get $\frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$.

(C) Given the partial fraction decomposition: $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.
Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$,we have $2x = A(x^2 + x + 1) + (Bx + C)(x - 1)$.
Setting $x = 1$: $2(1) = A(1^2 + 1 + 1) + 0 \implies 2 = 3A \implies A = \frac{2}{3}$.
Expanding the right side: $2x = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.
Equating coefficients of $x^2$: $A + B = 0 \implies B = -A = -\frac{2}{3}$.
Equating constant terms: $A - C = 0 \implies C = A = \frac{2}{3}$.
Thus,$A = \frac{2}{3}$,$B = -\frac{2}{3}$,and $C = \frac{2}{3}$.
Comparing these values,$A = C \neq B$.

(D) Let $\frac{x^2 + 1}{(2x - 1)(x - 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x - 1} + \frac{C}{x + 1}$.
Multiplying both sides by $(2x - 1)(x - 1)(x + 1)$,we get:
$x^2 + 1 = A(x - 1)(x + 1) + B(2x - 1)(x + 1) + C(2x - 1)(x - 1)$.
For $x = 1/2$: $(1/4) + 1 = A(1/2 - 1)(1/2 + 1) \implies 5/4 = A(-1/2)(3/2) \implies 5/4 = -3/4 A \implies A = -5/3$.
For $x = 1$: $1 + 1 = B(2(1) - 1)(1 + 1) \implies 2 = B(1)(2) \implies 2 = 2B \implies B = 1$.
For $x = -1$: $(-1)^2 + 1 = C(2(-1) - 1)(-1 - 1) \implies 2 = C(-3)(-2) \implies 2 = 6C \implies C = 1/3$.
Thus,the partial fraction is $\frac{-5}{3(2x - 1)} + \frac{1}{(x - 1)} + \frac{1}{3(x + 1)}$.

(D) We have the expression $\frac{2x}{x^4 + x^2 + 1}$.
First,factor the denominator: $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1)$.
Now,we express the fraction as partial fractions: $\frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{Ax + B}{x^2 - x + 1} + \frac{Cx + D}{x^2 + x + 1}$.
By partial fraction decomposition,we find $\frac{2x}{x^4 + x^2 + 1} = \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1}$.
Thus,the correct option is $D$.

(A) Given the expression $\frac{(x - a)(x - b)}{(x - c)(x - d)} = \frac{A}{x - c} - \frac{B}{x - d} + C$.
To find $C$,we perform polynomial long division on the left side.
The numerator is $(x - a)(x - b) = x^2 - (a + b)x + ab$.
The denominator is $(x - c)(x - d) = x^2 - (c + d)x + cd$.
Dividing $x^2 - (a + b)x + ab$ by $x^2 - (c + d)x + cd$:
$\frac{x^2 - (a + b)x + ab}{x^2 - (c + d)x + cd} = 1 + \frac{(c + d - a - b)x + (ab - cd)}{x^2 - (c + d)x + cd}$.
Comparing this with the given form $\frac{A}{x - c} - \frac{B}{x - d} + C$,we identify $C = 1$.