$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \mathrm{d} x=$
$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \mathrm{d} x=$
- $2 x \tan ^{-1} x-\log \left(1+x^2\right)+\mathrm{c}$, where c is a constant of integration.
- $2\left(x \tan ^{-1} x-\log \left(1+x^2\right)\right)+\mathrm{c}$, where c is a constant of integration.
- $x \tan ^{-1} x+\log \left(1+x^2\right)+\mathrm{c}$, where c is a . constant of integration.
- $2\left(x \tan ^{-1} x+\log \left(1+x^2\right)\right)+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{array}{ll} & I=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \\ & \text { Let } x=\tan t \\ \therefore \quad & d x=\sec ^2 t d t\end{array}$
$\begin{aligned} \therefore \quad \mathrm{I} & =\int \sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \sec ^2 t \mathrm{dt} \\ & =\int\left(\sin ^{-1}(\sin 2 \mathrm{t}) \sec ^2 \mathrm{t}\right) \mathrm{dt} \\ & =\int 2 \mathrm{t} \sec ^2 \mathrm{t} d \mathrm{t} \\ & =2\left[\mathrm{t} \int \sec ^2 \mathrm{t} d \mathrm{t}-\int \frac{\mathrm{dt}}{\mathrm{dt}}\left(\int \sec ^2 \mathrm{t} d \mathrm{dt}\right) \mathrm{dt}\right]+\mathrm{c} \\ & =2\left[\mathrm{t} \tan \mathrm{t}-\int \tan \mathrm{t} \mathrm{dt}\right]+\mathrm{c} \\ & =2 \mathrm{t} \tan \mathrm{t}+2 \log |\cos \mathrm{t}|+\mathrm{c} \\ & =2 x \tan ^{-1} x+2 \log \left|\frac{1}{\sqrt{1+x^2}}\right|+\mathrm{c} \\ & =2 x \tan ^{-1} x-\log \left(1+x^2\right)+\mathrm{c}\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 2)
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