solving $\int x^7\sqrt{3+2x^4}dx$

Question

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?

Answer 1

Try $u=x^4\to du=4x^3 dx$ instead, and note that $4x^7dx = udu$. Then: $$\int x^7\sqrt{3+2x^4}dx = \frac{1}{4}\int u\sqrt{3+2u}du$$ Now, it's easier to take $v=3+2u, dv=2du$ to get: $$\frac{1}{16}\int (v-3)\sqrt{v}dv$$ Which you can probably solve.

Answer 2

Using your substitution, we have $\displaystyle x=\left(\frac{u-3}{2}\right)^{1/4}$. Substitute this formula for $x$ into your new integral and solve!

Answer 3

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.