Question

${\int }_0^{\infty }\frac{x^2}{(x^2+4)(x^2+9)}dx$ is equal to ?

• A. $\frac{\pi }{20}$
• B. $\frac{\pi }{40}$
• C. $\frac{\pi }{10}$
• D. $\frac{\pi }{80}$

Answer 1

The correct option is C $\frac{\pi }{10}$

let $x^2=t$

${\int }_0^{\infty }\frac{x^2}{(x^2+y)(x^2+9)}\Rightarrow \frac{t}{(t+y)(t+9)}=\frac{A}{(t+y)}+\frac{B}{(t+9)}$

Comparing Coefficients on both sides

$t=A(t+9)+B(t+4)$

$t=(A+B)t+9A+4B$

$A+B=19A+4B=0$

$5A+4(A+B)=0$

$A=-4/5.B\frac{9}{5}$

${\int }_0^{\infty }-\frac{4}{5(t+4)}+\frac{9}{5(t+9)}\Rightarrow {\int }_0^{\infty }\frac{-4}{5(x^2+4)}+\frac{9}{5(x^2+9)}.dx$

$=\frac{-4}{5}{\int }_0^{\infty }\frac{1}{x^2+4}.dx+\frac{9}{5}{\int }_0^{\infty }\frac{1}{x^2+9}.dx$

$=\frac{-4}{5}\times \frac{1}{2}{\tan }^{-1}\frac{x}{2}+\frac{9}{5}\times \frac{1}{3}{\tan }^{-1}\frac{x}{3}+c$

$={[\frac{-2}{5}{\tan }^{-1}\frac{x}{5}+\frac{3}{5}{\tan }^{-1}\frac{x}{3}]}_0^{\infty }$

$=\frac{-2}{5}\times \frac{\pi }{2}+\frac{3}{5}\times \frac{\pi }{2}=\frac{\pi }{10}$