# 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}$$\int x^4\sqrt{u} {du} Since there is still a x variable in the integral, I'm not sure where to go from here. Any ideas? - You almost did it... write x^4 as (u-3)/2 in your last expression, you will get a primitive expressed by means of u, replace u by 3+2x^4 and you're done ! – Tom-Tom Feb 10 '14 at 9:28 ## 3 Answers Try u=x^4\to du=4x^3 dx instead, and note that 4x^7dx = udu. Then:$$\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. - 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! - 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$\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}which is \begin{align*} &\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}\\ =&\frac{1}{16}\int (u^{3/2}-3u^{1/2})du\\ =&\frac{1}{16}(\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2})+c\\ =&\frac{1}{16}(\frac{2}{5}(3+2x^4)^{5/2}-2(3+2x^4)^{3/2})+c \end{align*} wherec$is constant of integration. - 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