Chapter 13 - Section 13.2 - Substitution - Exercises - Page 971: 38

$\int\frac{x}{\sqrt {x+1}}dx=\frac{2}{3}(x+1)^{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}+C$

Substitution: $u=x+1$ $x=u-1$ $dx=du$ $\int\frac{x}{\sqrt {x+1}}dx=\int\frac{u-1}{u^{\frac{1}{2}}}du=\int(u^{\frac{1}{2}}-u^{-\frac{1}{2}})du=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{2}{3}(x+1)^{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}+C$