Q115 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Mathematics
The value of $\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}$ $\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$, for $x>0$ is
- Correct answerA$e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c$
- B$e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)+c$
- C$e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^3+c$
- D$-e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c$
Solution
Let $I=\int \frac{e^{\tan ^{-1} x}}{1+x^2}$
$$
\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x(x>0)
$$
Let us take $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta$
$$
\begin{aligned}
& \frac{1}{1+x^2} d x=d \theta \\
& \mathrm{I}=\int e^\theta\left[\left(\sec ^{-1}\left(\sqrt{1+\tan ^2 \theta}\right)\right)^2\right. \\
& \left.+\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\right] d \theta \\
& =\int e^\theta\left\{\left(\sec ^{-1} \sec \theta\right)^2+\left(\cos ^{-1} \cos 2 \theta\right)\right\} d \theta \\
& {\left[\because \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]} \\
& =\int e^\theta\left\{\theta^2+2 \theta\right\} d \theta\left[\because \sec ^{-1} \sec \theta=\theta\right. \text { and } \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \cos ^{-1} \cos 2 \theta=2 \theta] \\
& {\left[\because \int e^x\left(f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right] }
\end{aligned}
$$
So,
$$
\begin{aligned}
& I=e^\theta \cdot \theta^2+c \\
& I=e^{\tan ^{-1} x} \cdot\left(\tan ^{-1} x\right)^2+c
\end{aligned}
$$
Hence, the correct answer is option: 'A'.