<s>[INST] solving $\int x^7\sqrt{3+2x^4}dx$ I'm trying to solve $\int x^7\sqrt{3+2x^4}dx$
All I have so far is 
Let $u$ = $3+2x^4$
$du$ = $8x^3$ $dx$
$\frac{du}{8x^3}$ = $dx$
Therefore,
$\int x^7\sqrt{u}$ $\frac{du}{8x^3}$
$\frac{1}{8}$$\int x^4\sqrt{u}$ ${du}$
Since there is still a $x$ variable in the integral, I'm not sure where to go from here. Any ideas?
 [/INST] Given $\int x^7 \sqrt{3+2x^4} dx$. Assume that $3+2x^4= u$ then $8x^3 dx=du$. Hence the givenm integral becomes $\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}$ which is \begin{align*}
 &\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}\\
=&\frac{1}{16}\int (u^{3/2}-3u^{1/2})du\\
=&\frac{1}{16}(\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2})+c\\
=&\frac{1}{16}(\frac{2}{5}(3+2x^4)^{5/2}-2(3+2x^4)^{3/2})+c
\end{align*}
where $c$ is constant of integration.
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