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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

Your integral is

$$\int \frac{x \cdot x^2}{\sqrt[5]{x^2+3}}dx$$

So the natural choice is $u=x^2+3$.

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I have thought of u=x^2+3,too. but then I do not know how to express it in the numerator – Jaden Q Oct 2 '12 at 2:58
HINT: $u=x^2+3$ implies $x^2=u-3$, and $x dx = \frac{du}{2}$. – N. S. Oct 2 '12 at 2:59
oh yea, thanks so much for your help! i really do appreciate it – Jaden Q Oct 2 '12 at 3:02

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