Answer
$\frac{2}{3}(1+\sqrt{x})^3-3(1+\sqrt{x})^2+6(1+\sqrt{x})-2\ln (1+\sqrt{x})+C$
Work Step by Step
$\int \frac{x}{1+\sqrt{x}}dx$
Let $u=1+\sqrt{x}$.
Then,
$x=(u-1)^2$
and
$du=\frac{1}{2\sqrt{x}}dx$
$du=\frac{1}{2(u-1)}dx$
$dx={2(u-1)}\cdot du$
Using the $u-$substitution method,
$\int \frac{x}{1+\sqrt{x}}dx=\int \frac{(u-1)^2}{u}\cdot 2(u-1) \cdot du$
$=\int \frac{2(u-1)^3}{u} du$
$=\int \frac{2u^3-6u^2+6u-2}{u}du$
$=\int 2u^2-6u+6-\frac{2}{u}\ du$
$=\frac{2}{3}u^3-3u^2+6u-2\ln u+C$ (Back substitute $u=1+\sqrt{x}$
$=\frac{2}{3}(1+\sqrt{x})^3-3(1+\sqrt{x})^2+6(1+\sqrt{x})-2\ln (1+\sqrt{x})+C$
