Question

Resolve into partial fractions $\frac{4x^3-7x^2+5x-4}{(x^2+1)(x^2-3x+2)}$

• A. $\frac{x}{x^2+1}+\frac{1}{x-2}+\frac{x}{x-1}$
• B. $\frac{x}{x^2+1}+\frac{2}{x-2}+\frac{x}{x-1}$
• C. $\frac{x}{x^2+1}+\frac{2}{x-2}+\frac{1}{x-1}$
• D. $\frac{x}{x^2+1}+\frac{2}{x-2}+\frac{2x}{x-1}$

Answer 1

Answer: (C)

$\frac{x}{x^2+1}+\frac{2}{x-2}+\frac{1}{x-1}$

Let $\frac{4x^3-7x^2+5x-4}{(x^2+1)(x^2-3x+2)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-2}+\frac{D}{x-1}$
$\Rightarrow 4x^3-7x^2+5x-4=(Ax+B)(x-2)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x-2)\Rightarrow 4x^3-7x^2+5x-4=A(x^3-3x^2+2x)+B(x^2-3x+2)+C(x^3-x^2-x-1)+D(x^3-2x^2+x-2)$
On comparing coefficients we get
$A+C+D=4,-3A+B-C-2D=-7,2A-3B-C+D=5,2B-C-2D=-4\Rightarrow A=1,B=0,C=2,D=1$
Hence
$\frac{4x^3-7x^2+5x-4}{(x^2+1)(x^2-3x+2)}=\frac{x}{x^2+1}+\frac{2}{x-2}+\frac{1}{x-1}$
Hence, option 'C' is correct.