Q95 / 160 · TS EAMCET 2015 · Mathematics
If $\int e^{-x} \tan ^{-1}\left(e^x\right) d x$ $=f(x)-\frac{1}{2} \log \left(1+e^{2 x}\right)+C$, then $f(x)$ equals
- A$e^x-e^{-x} \tan ^{-1}\left(e^x\right)$
- B$x^2+e^{-x} \tan ^{-1}\left(e^x\right)$
- C$-e^{-x} \tan ^{-1}\left(e^x\right)$
- Correct answerD$x-e^{-x} \tan ^{-1}\left(e^x\right)$
Solution
Let $I=\int e^{-x} \tan ^{-1}\left(e^x\right) d x$
Put $e^x=t$
$$
\begin{aligned}
& \Rightarrow \quad e^x d x=d t \\
& \Rightarrow \quad d x=\frac{1}{e^x} d t \Rightarrow d x=\frac{1}{t} d t
\end{aligned}
$$
$$
\begin{gathered}
\therefore \quad I=\int \frac{\tan ^{-1} t}{t^2} d t \\
=-\frac{1}{t} \tan ^{-1} t-\int \frac{-1}{t} \cdot \frac{1}{1+t^2} d t
\end{gathered}
$$
[using integration by parts]
$$
\begin{aligned}
& =\frac{-1}{t} \tan ^{-1} t+\int\left(\frac{1}{t}-\frac{t}{1+t^2}\right) d t \\
& =\frac{-1}{t} \tan ^{-1} t+\log t-\frac{1}{2} \log \left(1+t^2\right)+C \\
& =-e^{-x} \tan ^{-1}\left(e^x\right)+\log \left(e^x\right)-\frac{1}{2} \log \left(1+e^{2 x}\right)+C \\
\therefore & f(x)=-e^{-x} \tan ^{-1}\left(e^x\right)+x=x-e^{-x} \tan ^{-1}\left(e^x\right)
\end{aligned}
$$
Hence, the correct answer is option: 'D'.