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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

1 Answers 1

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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 numerator2012-10-02
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    HINT: $u=x^2+3$ implies $x^2=u-3$, and $x dx = \frac{du}{2}$.2012-10-02
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    oh yea, thanks so much for your help! i really do appreciate it2012-10-02