# How to make a substitution for this integrand?

The integral is: $$\int \frac{x^3}{\sqrt[5]{x^2+3}} \mathrm{d}x$$ I am confused of which should be included in the substitution. Please help! Thank you so much!

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I would probably let $u^3=x^2+3$. Then $3u^2\,du=2x\,dx$, and when the smoke clears we are integrating $(3/2)(u^4-3u)$. –  André Nicolas Oct 2 '12 at 3:13

$$\int \frac{x \cdot x^2}{\sqrt[5]{x^2+3}}dx$$
So the natural choice is $u=x^2+3$.
HINT: $u=x^2+3$ implies $x^2=u-3$, and $x dx = \frac{du}{2}$. –  N. S. Oct 2 '12 at 2:59