$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=($ where $|x| < 1)$

$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=($ where $|x| < 1)$
  1. $2 \tan ^{-1} x-\log \left|1+x^2\right|+c$
  2. $x \tan ^{-1} x+\log \left|1+x^2\right|+c$
  3. $\tan ^{-1} x+\log \left|1+x^2\right|+c$
  4. $2 x \tan ^{-1} x-\log \left|1+x^2\right|+c$

Solution

Let $I=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$ When $x=\tan \theta, \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta$ and $\begin{aligned} & \mathrm{dx}=\sec ^2 \theta \mathrm{d} \theta \\ & \therefore \mathrm{I}=\int 2 \theta \sec ^2 \theta \mathrm{d} \theta \\ & =2 \int \theta \sec ^2 \theta d \theta=2\left[[\theta \tan \theta]-\int \tan \theta d \theta\right] \\ & =2[\theta \operatorname{tran} \theta+\log |\cos \theta|]+\mathrm{c} \\ & =2\left(\tan ^{-1} \mathrm{x}\right)(\mathrm{x})+2 \log \left|\sqrt{\frac{1}{1+\tan ^2 \theta}}\right|+\mathrm{c} \\ & =2 x \tan ^{-1} x+2 \log \left|\sqrt{\frac{1}{1+x^2}}\right|+c \\ & =2 x \tan ^{-1} x+2 \log \left|1+x^2\right|^{-\frac{1}{2}}+c=2 x \tan ^{-1} x-\log \left|1+x^2\right|+c \\ & \end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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