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
$$\int\frac{\cosh^3t}{\sinh^2t}dt$$
If you want to do this by using the substitution $x = \cosh(t)$ you also need $dx = \sinh(t)dt$, which means $$\int \frac{x^3}{\sqrt{x^2-1}}dx = \int \frac{\cosh^3(t)\sinh(t)dt}{\sqrt{\cosh^2(t)-1}}$$ and now use $\cosh^2(t) = 1+\sinh^2(t)$ in the numerator to get $$\int \frac{\cosh^3(t)\sinh(t)dt}{\sqrt{\cosh^2(t-11}} = \int \frac{(1+\sinh^2(t))\cosh(t)\sinh(t)dt}{\sqrt{\cosh^2(t)-1}} \\ =\int \frac{\cosh(t)\sinh(t)+\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt \\ = \int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt+\int \frac{\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt$$ The first half of that integral $\int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt$ is extremely easy if you make the substitution $u = \cosh^2(t)-1$. For the second integral, use integration by parts with $$u = \sinh^2(t), \quad du = 2\sinh(t)\cosh(t)dt, \quad dv = \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt$$ where $v$ is the first half of the integral you already solved. Hence, $$\int \frac{\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt = \int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}\sinh^2(t)dt \\ = uv-\int v du \\ = \sinh^2(t) \cdot (\text{Integral you already solved})-\int 2\sinh(t)\cosh(t)\cdot (\text{Integral you already solved})dt$$ For the new integral obtained through integration by parts, you should again be able to use $u =\cosh^2(t)-1$ for a pretty easy result.