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!
How to make a substitution for this integrand?
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$\begingroup$
calculus
integration
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0I 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
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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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0I have thought of u=x^2+3,too. but then I do not know how to express it in the numerator – 2012-10-02
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0HINT: $u=x^2+3$ implies $x^2=u-3$, and $x dx = \frac{du}{2}$. – 2012-10-02
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0oh yea, thanks so much for your help! i really do appreciate it – 2012-10-02