Question

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

• A. $2\pi$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{16}$
• D. $\frac{\pi }{32}$

Answer 1

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

Let

$I={\int }_0^{\infty }\frac{xdx}{(1+x)(x^2+1)}$
By partial fraction,
$\frac{x}{(1+x)(x^2+1)}=\frac{A}{(1+x)}+\frac{Bx+C}{(x^2+1)}$
$\Rightarrow x=A(x^2+1)+(1+x)(Bx+C)$
$\Rightarrow x=A(x^2+1)+(Bx+B{x}^2+C+Cx)$
$\Rightarrow x=(A+B)x^2+(B+C)x+(A+C)$

On comparing both sides, we get
$A+B=0,B+C=1,A+C=0$ ..... (i)

On adding all these equations, we get
$A+B+C=\frac{1}{2}$ ...... (ii)$\therefore A=\frac{1}{2}-1=-\frac{1}{2},C=\frac{1}{2}$

and $B=\frac{1}{2}$

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

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

$=-\frac{1}{2}\left[\log \right(1+x){]}_0^{\infty }+\frac{1}{4}[\log (x^2+1){]}_0^{\infty }+\frac{1}{2}\times \frac{\pi }{2}$

$=-\frac{1}{2}\underset{x\to \infty }{lim}\log (1+x)+\frac{1}{4}\underset{x\to \infty }{lim}\log (1+x^2)+\frac{\pi }{4}$

$=\underset{x\to \infty }{lim}\log \left[\frac{(1+x^2{)}^{1/4}}{(1+x{)}^{1/2}}\right]+\frac{\pi }{4}$

$=\underset{x\to \infty }{lim}\log \left[\frac{\sqrt{x}{(\frac{1}{x^2}+1)}^{1/4}}{\sqrt{x}{(\frac{1}{x}+1)}^{1/2}}\right]+\frac{\pi }{4}$

$=\log \frac{(0+1{)}^{1/4}}{(0+1{)}^{1/2}}+\frac{\pi }{4}$

$=\log \left(1\right)+\frac{\pi }{4}=0+\frac{\pi }{4}=\frac{\pi }{4}$