Mathematics

# Solve $\int \tan^{-1}\left(\dfrac{2x}{1-x^{2}}\right)dx$

##### SOLUTION
$\displaystyle\int \tan^{-1}\left(\dfrac{2x}{1-x^{2}}\right)dx$
Let $x=\tan\theta\Rightarrow \sec\theta=\sqrt{1+\tan^{2}\theta}=\sqrt{1+x^{2}}$
$\therefore \displaystyle\int \tan^{-1}\left(\dfrac{2\tan\theta}{1-\tan\theta}\right).\sec^{2}\theta d\theta$
$=\displaystyle\int \tan^{-1}(\tan 2\theta).\sec^{2}\theta\ d\theta$
$=\displaystyle\int 2\theta\sec^{2}\theta\ d\theta$
$=2\theta\tan\theta-2\displaystyle\int \tan\theta\ d\theta$
$=2\theta \tan\theta-2\ln |\sec\theta|+c$
$=2x\tan^{-1}x-2\ln |\sqrt{1+x^{2}}|+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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