Question

$\int \frac{(x-1)}{(x+1)(x^2+1)}dx=$

• A. $\frac{1}{2}\log |\frac{x+1}{\sqrt{x^2+1}}|+c$
• B. $-\log |x+1|+{\tan }^{-1}x+c$
• C. $\log |\frac{\sqrt{x^2+1}}{x+1}|+c$
• D. $\log |x+1|+\frac{1}{2}{\tan }^{-1}x+c$

Answer 1

Answer: (B)

$-\log |x+1|+{\tan }^{-1}x+c$

$\int \frac{(x-1)}{(x+1)(x^2+1)}dx$
$\frac{(x-1)}{(x+1)(x^2+1)}=\frac{A}{(x+1)}+\frac{Bx+C}{(x^2+1)}$
$\Rightarrow x-1=A(x^2+1)+(Bx+C)(x+1)$
$x-1=(B+C)x+(A+B)x^2+(A+C)$
$\Rightarrow B+C=1,A+B=0,A+C=-1$
Solving these equations, we get
$C=0,A=-1,B=1$
$\frac{(x-1)}{(x+1)(x^2+1)}=\frac{-1}{(x+1)}+\frac{1}{(x^2+1)}$
Integrating w.r.t x
$\int \frac{(x-1)}{(x+1)(x^2+1)}dx=$ $-\log |x+1|+{\tan }^{-1}x+c$