Q94 / 160 · TS EAMCET 2015 · Mathematics
$\int \sqrt{e^x-4} d x$ equals
- A$\tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+\sqrt{e^x-4}+C$
- Correct answerB$2 \sqrt{e^x-4}-4 \tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C$
- C$2 \sqrt{e^x-4}-4 \cot ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C$
- D$\sqrt{e^x-4}-4 \tan ^{-1}\left(\sqrt{e^x-4}\right)+C$
Solution
$$
\begin{aligned}
& \text { Let } I=\int \sqrt{e^x-4} d x \\
& \begin{aligned}
\text { Put } e^x-4 & =t^2 \\
\Rightarrow \quad e^x d x & =2 t d t \\
\Rightarrow \quad d x & =\frac{2 t}{t^2+4} d t \\
\therefore \quad I & =\int t \cdot \frac{2 t}{t^2+4} d t \\
& =2 \int \frac{t^2}{t^2+4} d t=2 \int \frac{t^2+4-4}{t^2+4} d t \\
& =2\left[\int 1-\frac{4}{t^2+(2)^2} d t\right] \\
& =2\left[t-4 \cdot \frac{1}{2} \tan ^{-1} \frac{t}{2}\right]+C \\
& =2 t-4 \tan ^{-1} \frac{t}{2}+C \\
& =2 \sqrt{e^x-4}-4 \tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C
\end{aligned}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.