Answer
The answer is $\int \frac{x}{\sqrt{x+1}}
dx = 2x (x+1)^{\frac{1}{2}}
- \frac{4}{3} (x+1)^{\frac{3}{2}} + c$.
Work Step by Step
$\int \frac{x}{\sqrt{x+1}} dx = \int x (x+1)^{\frac{-1}{2}} dx$, and
for $u=x$, then $du = dx$ and for $dv= (x+1)^{\frac{-1}{2}} dx$,
then $v = 2 (x+1)^{\frac{1}{2}}$. Hence, integration by parts $\int
u dv = u v - \int v du$ implies that $\int \frac{x}{\sqrt{x+1}}
dx = x \times 2 (x+1)^{\frac{1}{2}}
- \int 2 (x+1)^{\frac{1}{2}} dx = 2x (x+1)^{\frac{1}{2}}
- \frac{4}{3} (x+1)^{\frac{3}{2}} + c$.
