Question

$\stackrel{\infty }{\underset{0}{\int }}\frac{x}{(1+x)(1+x^2)}dx$ equals to:

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. Is same as $\stackrel{\infty }{\underset{0}{\int }}\frac{dx}{(1+x)(1+x^2)}dx$
• D. Cannot be evaluated

Answer 1

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

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

$\frac{x}{(1+x)(1+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{(x^2+1)}$

$\frac{x}{(1+x)(1+x^2)}=\frac{A(x^2+1)+Bx(x+1)+C(x+1)}{(1-x)(x^2+1)}$

$x=A(x^2+1)+Bx(x+1)+C(x+1)$

Put $x=-1$

$-1=2A$

$A=\frac{-1}{2}$

Put $x=0$

$0=\frac{-1}{2}+C$

$C=1/2$

Put $x=1$

$1=-1+B\left(2\right)+1$

$B=1/2$

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

$=\frac{1}{2}[-\ln (x+1)+{\tan }^{-1}x+\frac{1}{2}\ln (x^2+1\left)\right]{|}_0^{\infty }$

$=\frac{1}{2}[\ln \left(\frac{\sqrt{x^2}+1}{x+1}\right)+{\tan }^{-1}x]{|}_0^{\infty }$

$=\frac{1}{2}.\frac{\pi }{2}$

$=\frac{\pi }{2}$

${\int }_0^{\infty }\frac{xdx}{(1+x)(1+x^2)}=\frac{\pi }{4}$