How to use integration by parts to solve an integral?

Question

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$ $dt = 2x dx$

$v_1$ = $\frac{1}{2}\int \sqrt{t}dt$

$v_1$ = $\frac{1}{2}(\frac{2}{3}t^\frac{3}{2}) + c$

$v_1$ = $\frac{1}{3} (x^2-1)^\frac{3}{2} + c$

$uv - \int vdu$

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

$u_2 = 2x$ $dv_2 = \frac{1}{3} (x^2-1)^\frac{3}{2}dx$

$du_2 = 2dx$

Answer 1

I would make the substitution $u=x^2-1$ first. This makes the integral

$$\frac{1}{2}\int (u-1)\sqrt{u} \; du.$$

Parts will work on this integral, but it's not the easiest way.

Answer 2

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*}

Answer 3

You used integration by parts once. And you used substitution to find $v_1$. Use substitution again: $$I=\int 2x\cdot \frac13(x^2-1)^{3/2}dx;$$ $$t=x^2-1; \ \ dt=2xdx;$$ $$I=\frac13 \int t^{3/2}dt=\frac13\cdot \frac25 \cdot t^{5/2}+C=\frac{2}{15}(x^2-1)^{5/2}+C.$$