Question

${\int }_0^{\infty }\frac{x}{(1+x)(1+x^2)}dx=$

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{4}$

${\int }_0^{\infty }\frac{x}{(1+x)(1+x^2)}dx$
Let $I=\int \frac{x}{(1+x)(1+x^2)}dx$
$\frac{x}{(1+x)(1+x^2)}=\frac{A}{x+1}+\frac{Bx+C}{1+x^2}$
$\Rightarrow x=A(1+x^2)+(Bx+C)(1+x)$
$\Rightarrow A+B=0$
$B+C=1$
$A+C=0$
On solving , we get $A=-\frac{1}{2},B=\frac{1}{2},C=\frac{1}{2}$
So, $I=-\frac{1}{2}\int \frac{1}{1+x}dx+\frac{1}{2}\int \frac{x+1}{x^2+1}dx$
$=-\frac{1}{2}\log |1+x|+\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx$
$I=-\frac{1}{2}\log |1+x|+\frac{1}{4}\int \frac{1}{t}dt+\frac{1}{2}{\tan }^{-1}x$
$I=-\frac{1}{2}\log |1+x|+\frac{1}{4}\log |1+x^2|+\frac{1}{2}{\tan }^{-1}x$
So, ${\int }_0^{\infty }\frac{x}{(1+x)(1+x^2)}dx=\frac{1}{2}\left(\frac{\pi }{2}\right)=\frac{\pi }{4}$