A level Integration: $\int\frac{x^3}{\sqrt{x^2-1}}dx$ Using the substitution $x=\cosh (t)$ or otherwise, find 
$$\int\frac{x^3}{\sqrt{x^2-1}}dx$$
The correct answer is apparently 
$$\frac{1}{3}\sqrt{x^2-1}(x^2+2)$$
I seem to have gone very wrong somewhere; my answer is way off, can someone explain how to get this answer to me.
Thanks.
My working:
$$\int\frac{\cosh^3t}{\sinh^2t}dt$$
$$u=\sinh t$$
$$\int\frac{1+u^2}{u^2}du$$
$$\frac{-1}{u}+u$$
$$\frac{-1}{\sinh t}+\sinh t$$
$$\frac{-1}{\sqrt{x^2-1}}+\sqrt{x^2-1}$$
^my working, I'm pretty sure this is very wrong though.
Edit: I've spotted my error. On the first line it should be 
$$\int \cosh^3t \, dt$$ 
not
$$\int\frac{\cosh^3t}{\sinh^2t}dt$$

Let $\displaystyle u=x^2, dv=\frac{x}{\sqrt{x^2-1}}dx$  $\;\;$and $\;\;du=2xdx, v=\sqrt{x^2-1}$ to get
$\displaystyle\int\frac{x^3}{\sqrt{x^2-1}}\;dx=x^2\sqrt{x^2-1}-\int2x\sqrt{x^2-1}\;dx=x^2\sqrt{x^2-1}-\frac{2}{3}(x^2-1)^{3/2}+C$

Alternatively, let $u=\sqrt{x^2-1}, \;\;du\displaystyle=\frac{x}{\sqrt{x^2-1}}dx, \;\;x^2=u^2+1$ to get
$\displaystyle\int\left(u^2+1\right)du=\frac{1}{3}u^3+u+C=\frac{1}{3}(x^2-1)^{3/2}+\sqrt{x^2-1}+C$
