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If $\quad \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$ then $f(x)$ is equal to
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The correct answer is:
$2 x \tan ^{-1} x$
Let $I=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\begin{aligned}
\therefore \quad I & =\int \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta d \theta \\
& =2 \int \theta \sec ^2 \theta d \theta \\
& =2\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =2[\theta \tan \theta+\log \cos \theta]+c \\
& =2\left[x \tan ^{-1} x+\log \frac{1}{\sqrt{1+x^2}}\right]+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c
\end{aligned}$
But $I=f(x)-\log \left(1+x^2\right)+c$
$\begin{array}{cc}
\Rightarrow & f(x)-\log \left(1+x^2\right)+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c \\
& f(x)=2 x \tan ^{-1} x
\end{array}$
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\begin{aligned}
\therefore \quad I & =\int \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta d \theta \\
& =2 \int \theta \sec ^2 \theta d \theta \\
& =2\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =2[\theta \tan \theta+\log \cos \theta]+c \\
& =2\left[x \tan ^{-1} x+\log \frac{1}{\sqrt{1+x^2}}\right]+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c
\end{aligned}$
But $I=f(x)-\log \left(1+x^2\right)+c$
$\begin{array}{cc}
\Rightarrow & f(x)-\log \left(1+x^2\right)+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c \\
& f(x)=2 x \tan ^{-1} x
\end{array}$
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