Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{d x}{\sqrt{1+x}+\sqrt{x}}=$
Options:
Solution:
1649 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{3}(1+x)^{3 / 2}-\frac{2}{3} x^{3 / 2}+c$
$\begin{aligned} & \int \frac{d x}{\sqrt{1+x}+\sqrt{x}}=\int\left[\frac{(x+1)-x}{\sqrt{1+x}+\sqrt{x}}\right] d x \\ & \int(\sqrt{x+1}-\sqrt{x}) d x=\frac{(x+1)^{3 / 2}}{3 / 2}-\frac{x^{3 / 2}}{3 / 2}+c \\ & =\frac{2}{3}\left[(x+1)^{3 / 2}-x^{3 / 2}\right]+c=\frac{2}{3}(x+1)^{3 / 2}-\frac{2}{3} x^{3 / 2}+c .\end{aligned}$
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.