Answer
$$\int\frac{1}{\sqrt x(1+\sqrt x)^2}dx=-\frac{2}{1+\sqrt x}+C$$
Work Step by Step
$$A=\int\frac{1}{\sqrt x(1+\sqrt x)^2}dx$$
We set $u=1+\sqrt x$.
Then $$du=(0+\frac{1}{2\sqrt x})dx=\frac{dx}{2\sqrt x}$$
That means, $$\frac{dx}{\sqrt x}=2du$$
Therefore, $$A=2\int\frac{1}{u^2}du=-2\int-\frac{1}{u^2}du$$ $$A=-2\times\frac{1}{u}+C=-\frac{2}{u}+C$$ $$A=-\frac{2}{1+\sqrt x}+C$$
