Integral $\int{\frac{x^5}{\sqrt{x^2+7}}}\,\text{d}x$. I am trying to solve this integral: $$\int{\frac{x^5}{\sqrt{x^2+7}}}\,\text{d}x.$$
I tried using the transform $x=\sqrt{7}\sinh{t}$ which leads to 
 $$\int{(\sinh{t}})^5\,\text{d}t=\int \left(\frac{\cosh{2t}-1}{2}\right)^2 \times\sinh{t}\,\text{d}t.$$
There I can find the integrals except $\int\cosh[2 t]^2\times \sinh[t]\,\text{d}t$. How can I find this?
 A: $$\int\frac{x^5}{\sqrt{x^2+7}}~dx~=~\int\frac{\big(x^2\big)^2~x}{\sqrt{x^2+7}}~dx~=~\frac12\int\frac{\Big[\big(x^2+7\big)-7\Big]^2}{\sqrt{x^2+7}}~d\big(x^2+7\big)=\int\frac{\big(t-7\big)^2}{2\sqrt t}~dt$$
Now let $t=u^2$, and use partial fraction decomposition. $\Big($Alternately, let $x=\sqrt7\cdot\sinh y\Big)$.
A: Hint: 
$$
  \int\text{sinh}^5t\:\text{d}t=\int(\text{sinh}^2t)^2 \text{sinh}\:t \:\text{d}t = \int(\text{cosh}^2 t-1)^2 \:\text{d}(\text{cosh}\:t).
$$
A: $$\int{\frac{x^5}{\sqrt{x^2+7}}}dx$$
Let $u = x^2 + 7 \implies du = 2x\,dx$ and $x^2 = u-7$. 
Then $$\begin{align} \int{\frac{x^5}{\sqrt{x^2+7}}}dx & = \frac 12\int\frac {(x^2)^2 \,(2x\,dx)}{(x^2 + 7)^{1/2}} \\ \\ & = \frac 12\int \frac{(u-7)^2 \,du}{u^{1/2}}\\ \\ & = \frac 12\int\frac{u^2 - 14u + 49}{u^{1/2}}\,du \\ \\ 
&= \frac 12 \int \left(u^{3/2} - u^{1/2} + 49 u^{-1/2}\right)\,du\\ \\ 
&= \frac 12\left( \frac 25u^{5/2} - \frac 23 u^{3/2} + 2u^{1/2}\right) +C\\ \\
&= \frac 15(x^2-7)^{5/2} - \frac 13(x^2 - 7)^{3/2} + (x^2-7)^{1/2} +C\\ \\
&= \frac 1{15} \sqrt{x^2-7}\left(3(x^2-7)^2 - 5(x^2 -7) + 15\right)  + C\\ \\
\end{align}$$
Of course, the last line can be simplified further.
A: take x as $\sqrt{7}\tan t$ and then proceed and second substitution take $\sec t=a$ these both should be enough 
A: The comment of André Nicolas has anticipated my answer.....
The substitution:
$$
x^2+7=t
$$
gives:
$$
2xdx=dt
$$
so we have:
$$
\int{\dfrac{x^5}{\sqrt{x^2+7}}} dx=
\int{\dfrac{x^5}{2x\sqrt{t}}} dt=
\dfrac{1}{2}\int{\dfrac{(t-7)^2}{\sqrt{t}}} dt
$$
that can be easily solved as a sum of powers.
If you want use your substitution note that:
$$
\sinh^5(t)=\sinh t (\cosh^2 t-1)^2 
$$
and the substitution:
$$
\cosh t = u
$$
gives
$$
\sinh t dt =du
$$
and you have again a polynomial....
