Question

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

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. is sme as ${\int }_0^{\infty }\frac{dx}{(1+x)(1+x^2)}$
• D. cannot be evaluated

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\pi }{4}$
C is sme as ${\int }_0^{\infty }\frac{dx}{(1+x)(1+x^2)}$

Let $I={\int }_0^{\infty }\frac{xdx}{(1+x)(1+x^2)}$
Using partial fraction
$={\int }_0^{\infty }(\frac{x+1}{2(1+x^2)}-\frac{1}{2(1+x)})dx={\left({lim}_{b\to \infty }\frac{1}{2}\log (1+x^2)-\frac{1}{2}\log (1+x)+\frac{1}{2}{tan}^{-1}x\right)}_0^b={lim}_{b\to \infty }(\frac{1}{2}\log (1+b)-\frac{1}{2}\log (1+b)+\frac{1}{2}{tan}^{-1}b)=\frac{\pi }{4}$