Answer
$$\int \frac{d x}{(1+\sqrt{x})^{2}}=2\left[\ln (1+\sqrt{x})+\frac{1}{1+\sqrt{x}}\right]+C$$
Work Step by Step
Given $$\int \frac{d x}{(1+\sqrt{x})^{2}}$$
So,
\begin{array}{l}{u=1+\sqrt{x}, u-1=\sqrt{x}} \\ {u=1+x^{1 / 2}} \\ {d u=\frac{1}{2} x^{-1 / 2} d x} \\ {d u=\frac{1}{2 \sqrt{x}} d x} \\ {d x=2 \sqrt{x} d u} \\ {d x=2(u-1) d u}\end{array}
So, we get
\begin{array}{l}{I=\int \frac{d x}{(1+\sqrt{x})^{2}}}\\{=2 \int \frac{u-1}{u^{2}} d u} \\ {=2 \int \frac{u}{u^{2}}-\frac{1}{u^{2}} d u} \\ {=2 \int \frac{1}{u}-\frac{1}{u^{2}} d u} \\ {=2\left(\ln u+\frac{1}{u}\right)+C} \\ {=2\left[\ln (1+\sqrt{x})+\frac{1}{1+\sqrt{x}}\right]+C}\end{array}
