Question

Resolve into partial fraction $\frac{7+x}{(1+x)(1+x^2)}$

• A. $\frac{5}{1+x}+\frac{4-3x}{1+x^2}$
• B. $\frac{3}{1+x}-\frac{4-3x}{1+x^2}$
• C. $\frac{3}{1+x}+\frac{4-3x}{1+x^2}$
• D. $\frac{5}{1+x}-\frac{4-3x}{1+x^2}$

Answer 1

Answer: (C)

$\frac{3}{1+x}+\frac{4-3x}{1+x^2}$

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