Question

${\int }^{}\frac{1}{\sqrt{x}(1+x)}dx$

Answer 1

We have,

$\int \frac{1}{\sqrt{x}(1+x)}dx$

Let

$t=\sqrt{x}$

$t^2=x$

$dt=\frac{1}{2\sqrt{x}}dx$

$dt=\frac{1}{2t}dx$

$dx=2tdt$

$\int \frac{1}{t(1+t^2)}dx$

$=\int \frac{1}{t(1+t^2)}2tdt$

$=\int \frac{1}{(1+t^2)}2dt$

$=2\int \frac{1}{1+t^2}dt$

$=2{\tan }^{-1}t+C$

$=2{\tan }^{-1}\left(\sqrt{x}\right)+C$

Hence, this is the answer.