Question

${\int }_0^1\frac{1}{(x^2+16)(x^2+25)}dx$ is equal to

• A. $\frac{1}{5}[\frac{1}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$
• B. $\frac{1}{9}[\frac{1}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$
• C. $\frac{1}{4}[\frac{1}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$
• D. $\frac{1}{9}[\frac{1}{5}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{4}{\tan }^{-1}\left(\frac{1}{5}\right)]$
• E. $\frac{1}{9}[\frac{3}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{4}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$

Answer 1

The correct option is B $\frac{1}{9}[\frac{1}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$

Let $I={\int }_0^1\frac{1}{(x^2+16)(x^2+25)}dx$
$=\frac{1}{9}{\int }_0^1(\frac{1}{x^2+16}-\frac{1}{x^2+25})dx$
$=\frac{1}{9}{[\frac{1}{4}{\tan }^{-1}\frac{x}{4}-\frac{1}{5}{\tan }^{-1}\frac{x}{5}]}_0^1$
$=\frac{1}{9}[\frac{1}{4}{\tan }^{-1}\left(\frac{1}{4}\right)-\frac{1}{5}{\tan }^{-1}\left(\frac{1}{5}\right)]$