Q115 / 160 · TS EAMCET 2005 · Mathematics
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
- Correct answerA$2 x \tan ^{-1} x$
- B$-2 x \tan ^{-1} x$
- C$x \tan ^{-1} x$
- D$-x \tan ^{-1} x$
Solution
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 $I=f(x)-\log \left(1+x^2\right)+c$
$\begin{array}{cc}
\Rightarrow & f(x)-\log \left(1+x^2\right)+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c \\
& f(x)=2 x \tan ^{-1} x
\end{array}$
Hence, the correct answer is option: 'A'.