Question

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

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

Answer 1

Answer: (B)

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