<s>[INST] How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t = x^2-1$ $dt = 2x dx$
$v_1$ = $\frac{1}{2}\int \sqrt{t}dt$
$v_1$ = $\frac{1}{2}(\frac{2}{3}t^\frac{3}{2}) + c$
$v_1$ = $\frac{1}{3} (x^2-1)^\frac{3}{2} + c$
$uv - \int vdu$
= $\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \int 2x\frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$u_2 = 2x$ $dv_2 = \frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$du_2 = 2dx$ 
 [/INST] You don't have to use integration by parts for $\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx$. 
$$\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx=\frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1)$$
So the integral is
\begin{align*}
\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1)&=\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \frac{2}{15}(x^2-1)^\frac{5}{2}+C\\
&=\frac{1}{15}(x^2-1)^\frac{3}{2}[5x^2-2(x^2-1)]+C\\
&=\frac{1}{15}(x^2-1)^\frac{3}{2}(3x^2+2)+C\\
\end{align*}
</s>