Question

Evaluate ${\int }_0^{\infty }\frac{x^2+1}{x^4+7x^2+1}dx$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{3}$

${\int }_0^{\infty }\frac{x^2+1}{x^4+7x^2+1}dx$

Let

$I={\int }_0^{\infty }\frac{x^2+1}{x^4+7x^2+1}dx$

$={\int }_0^{\infty }\frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}}$

$={\int }_0^{\infty }\frac{1+\frac{1}{x^2}}{x^2-2+\frac{1}{x^2}+9}$

$={\int }_0^{\infty }\frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+(3{)}^2}$

Let $x-\frac{1}{x}=t$

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

$I={\int }_0^{\infty }\frac{dt}{t^2+(3{)}^2}$

$I={[\frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+c]}_0^{\infty }$

$I={[\frac{1}{3}{\tan }^{-1}\left(\frac{x^2-1}{3x}\right)+c]}_0^{\infty }$

$I=\frac{1}{3}[\frac{\pi }{2}-\left(\frac{-\pi }{2}\right)]$

$I=\frac{\pi }{3}$