Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{x d x}{2(1+x)^{3 / 2}}$ is equal to
Options:
Solution:
2471 Upvotes
Verified Answer
The correct answer is:
$\frac{2+x}{\sqrt{1+x}}+C$
Let $I=\int \frac{x d x}{2(1+x)^{3 / 2}}$
On putting, $1+x=t$, we get $d x=d t$
$\begin{aligned}
\therefore \quad I & =\int \frac{(t-1) d t}{2 t^{3 / 2}}=\frac{1}{2}\left[\int t^{-1 / 2} d t-\int t^{-3 / 2} d t\right] \\
& =\frac{1}{2}\left[\frac{t^{1 / 2}}{1 / 2}-\frac{t^{-1 / 2}}{-1 / 2}\right]+C \\
& =\frac{1}{2} \times 2\left[\sqrt{t}+\frac{1}{\sqrt{t}}\right]+C \\
& =\frac{t+1}{\sqrt{t}}+C=\frac{x+2}{\sqrt{1+x}}+C
\end{aligned}$
On putting, $1+x=t$, we get $d x=d t$
$\begin{aligned}
\therefore \quad I & =\int \frac{(t-1) d t}{2 t^{3 / 2}}=\frac{1}{2}\left[\int t^{-1 / 2} d t-\int t^{-3 / 2} d t\right] \\
& =\frac{1}{2}\left[\frac{t^{1 / 2}}{1 / 2}-\frac{t^{-1 / 2}}{-1 / 2}\right]+C \\
& =\frac{1}{2} \times 2\left[\sqrt{t}+\frac{1}{\sqrt{t}}\right]+C \\
& =\frac{t+1}{\sqrt{t}}+C=\frac{x+2}{\sqrt{1+x}}+C
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.