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I'm trying to solve $\int x^7\sqrt{3+2x^4}dx$

All I have so far is

Let $u$ = $3+2x^4$

$du$ = $8x^3$ $dx$

$\frac{du}{8x^3}$ = $dx$


$\int x^7\sqrt{u}$ $\frac{du}{8x^3}$

$\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?

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You almost did it... write $x^4$ as $(u-3)/2$ in your last expression, you will get a primitive expressed by means of $u$, replace $u$ by $3+2x^4$ and you're done ! – Tom-Tom Feb 10 '14 at 9:28
up vote 3 down vote accepted

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.

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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!

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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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That's the general idea. You unfortunately made many mistakes in your calculation, and the final result is wrong. – Tom-Tom Feb 10 '14 at 9:31
@ V.Rossetto I would be happy to know my mistakes so that in future it won't happen again. Would you please let me know what are they ? – Anjan3 Feb 14 '14 at 5:38
You've corrected most of your mistakes in your last edit. It remains only one in the exponent $5/4$ in the final result: it should be $5/2$. – Tom-Tom Feb 14 '14 at 8:26
@V.Rossetto Opps !!! That was really indeed a mistake. Thank you so much for alerting me. I have done the editing. And also I apologise for such typing mistake. – Anjan3 Feb 17 '14 at 4:54

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