Question

$\int_{0}^{1} \frac{1}{\left(x^{2}+16\right)\left(x^{2}+25\right)} \,dx$ is equal to

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

Answer 1

Answer: (B)

$\frac{1}{9} \left[\frac{1}{4} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} tan^{-1}\left(\frac{1}{5}\right)\right]$

Answer 2

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

$I =\int_{0}^{1} \frac{1}{\left(x^{2}+16\right)\left(x^{2}+25\right)} d x$

$=\frac{1}{9} \int_{0}^{1}\left(\frac{1}{x^{2}+16}-\frac{1}{x^{2}+25}\right) d x$

$=\frac{1}{9}\left[\frac{1}{4} \tan ^{-1} \frac{x}{4}-\frac{1}{5} \tan ^{-1} \frac{x}{5}\right]_{0}^{1}$

$=\frac{1}{9}\left[\frac{1}{4} \tan ^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} \tan ^{-1}\left(\frac{1}{5}\right)\right]$