Integration by multiple substitutions I have the following integration.  The only instructions are to integrade by substitution.
(again I am sorry but I have no idea how to use MathJax.  I am going to have to sit down and figure it out because I am sure it would be easier for everyone.)
$$
\int x^5 \sqrt {x^3 - 8}\,dx  
$$
So far I have $u = x^3$,  $du = 3x^2dx$  giving me:
$$
\frac13 \int u \sqrt {u - 8}\,du  
$$
And this is where I am stuck.
Any help in the right direction would be fantastic.
Thank you
 A: In general, when you have an integral of the form
$$
\int R(x,\sqrt{ax+b})dx
$$
the substitution $t=\sqrt{ax+b}$ gives
$$
\int R\left(\frac{t^2-b}{a},t\right)\frac{2tdt}{a}
$$
If $R$ is a rational function (of two variables), the resulting integral is a rational one.
A: Hint. One may write
$$
\begin{align}
\int u\cdot \sqrt{u-8}\:du&=\int (u-8+8)\cdot \sqrt{u-8}\:du
\\&=\int (u-8)^{3/2}\:du+8\int (u-8)^{1/2}\:du
\end{align}
$$ then just substitute $v=u-8$ to get
$$
\int u\cdot \sqrt{u-8}\:du=\int v^{3/2}\:dv+8\int v^{1/2}\:dv
$$ the last terms are now easier to evaluate.
A: $$\displaystyle\dfrac{1}{3}\int u\sqrt{u-8}\,du$$
Now,
$$\displaystyle\dfrac{1}{3}\left[\int (u-8+8)\sqrt{u-8}\,du\right]$$
$$=\displaystyle\dfrac{1}{3}\left[\int(u-8)^{\frac{3}{2}}du+8\int(u-8)^{\frac{1}{2}}du\right]$$
$Let (u-8)=v$
$\Rightarrow\;du=dv$
$$\Rightarrow\;\displaystyle\dfrac{1}{3}\int\left[v^{\frac{3}{2}}dv+8\int v^{\frac{1}{2}}du\right]$$
Integrate resulting integral
$$=\dfrac{1}{3}\left[\dfrac{2(u-8)^{\frac{5}{2}}}{5}+\dfrac{8×2}{3}(u-8)^{\frac{3}{2}}\right]$$
Since you let $u=x^3$
$$=\dfrac{2}{15}(x^3-8)^{\frac{5}{2}}+\dfrac{16}{9}(x^3-8)^{\frac{3}{2}}$$
$$=\dfrac{6}{45}\left[(x^3-8)^\frac{5}{2}\right]+\dfrac{16×5}{9×5}(x^3-8)^{\frac{3}{2}}$$
$$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}\left[3(x^3-8)+40\right]$$
$$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}\left[3x^3-24+40\right]$$
$$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}(3x^3+16)+C$$
