University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{x}{\sqrt{1+x}}dx=2\Big(\frac{(\sqrt{1+x})^3}{3}-\sqrt{1+x}\Big)+C$$
$$A=\int\frac{x}{\sqrt{1+x}}dx$$ We set $u=\sqrt{1+x}$, which means $x+1=u^2$ and $x=u^2-1$ Then $$du=\frac{(1+x)'}{2\sqrt{1+x}}dx=\frac{1}{2\sqrt{1+x}}dx$$ That means $$\frac{1}{\sqrt{1+x}}dx=2du$$ Therefore, $$A=2\int (u^2-1)du$$ $$A=2\Big(\frac{u^3}{3}-u\Big)+C$$ $$A=2\Big(\frac{(\sqrt{1+x})^3}{3}-\sqrt{1+x}\Big)+C$$