## shayanreloaded Group Title Find the value of the integral one year ago one year ago

1. OpenSessame

http://www.webmath.com/nintegrate.html Best website that gives you step by step instructions!

2. zzr0ck3r

$\int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx$?

yes

4. cinar

$\Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x^3(\frac{1}{x^2}-1)}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{x\sqrt[3]{\frac{1}{x^2}-1}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{\frac{1}{x^2}-1}}{x^3}dx\\ \Large u=\frac{1}{x^2}=>du=\frac{-2}{x^3}dx\\ \Large \frac{-1}{2}\int\sqrt[3]{u-1}du\\$the reset is yours (: you can do one more subt.

5. cinar

I guess this subt. is better$\Large u=\frac{1}{x^2}-1=>du=\frac{-2}{x^3}dx\\ \Large \frac{-1}{2}\int\sqrt[3]{u}du\\ \Large \frac{-1}{2}\frac{3}{4}u^{\frac{4}{3}}\\ \Large \frac{-3}{8}(\frac{1}{x^2}-1)^{\frac{4}{3}}|_{\frac{1}3}^{1}=6\\$