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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 $(3/2)(u^4-3u)$.2012-10-02

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