Question

Solve $\underset{1}{\overset{-1}{\int }}\frac{d}{dx}ta{n}^{-1}\left(\frac{1}{x}\right)dx$

Answer 1

Consider the given integral.

$I={\int }_1^{-1}\frac{d}{dx}\left({\tan }^{-1}\left(\frac{1}{x}\right)\right)dx$

$I={\int }_1^{-1}\frac{1}{1+\frac{1}{x^2}}dx\left[\frac{d}{dx}({\tan }^{-1}x=\frac{1}{1+x^2})\right]$

$I={\int }_1^{-1}\frac{x^2}{x^2+1}dx$

$I={\int }_1^{-1}\frac{x^2+1-1}{x^2+1}dx$

$I={\int }_1^{-1}1dx-{\int }_1^{-1}\frac{1}{x^2+1}dx$

$I={\left[x\right]}_1^{-1}-{\left[{\tan }^{-1}\left(x\right)\right]}_1^{-1}$

$I=[-1-\left(1\right)]-[{\tan }^{-1}(-1)-{\tan }^{-1}\left(1\right)]$

$I=[-2]-[{\tan }^{-1}\left(\tan \frac{3\pi }{4}\right)-{\tan }^{-1}\left(\frac{\pi }{4}\right)]$

$I=[-2]-[\frac{3\pi }{4}-\frac{\pi }{4}]$

$I=[-2]-\left[\frac{\pi }{2}\right]$

$I=-2-\frac{\pi }{2}$

Hence, this is the answer.