Question

Integrate:
${\tan }^{-1}\left(\frac{x-a}{x+a}\right)$

Answer 1

$ta{n}^{-1}\left(\right(\frac{x-a}{x+a}\left)\right)=ta{n}^{-1}\left(\right(\frac{\left(\frac{x}{a}\right)-1}{\left(\frac{x}{a}\right)+1}\left)\right)let\left(\frac{x}{a}\right)=tan\theta or\theta =ta{n}^{-1}\left(\frac{x}{a}\right)=ta{n}^{-1}\left(\right(\frac{tan\theta -tan\left(\frac{\pi }{4}\right)}{1+tan\theta \times \tan \left(\frac{\pi }{4}\right)}\left)\right)=ta{n}^{-1}\left[tan(\theta -(\frac{\pi }{4}\left)\right)\right]=\theta -\left(\frac{\pi }{4}\right)=ta{n}^{-1}\left(\frac{x}{a}\right)-\left(\frac{\pi }{4}\right)$