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Math Help - some help with these integrals

  1. #1
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    some help with these integrals

    ok I'm having problems with these if anyone can help...


    integral of x^2(1+x^3)^(-1/3)dx



    and this other one

    integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

    This ends up in form alog3+blog2 I just need to know how to get there..
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    Quote Originally Posted by dankelly07 View Post
    ok I'm having problems with these if anyone can help...


    integral of x^2(1+x^3)^(-1/3)dx
    \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x

    Substituting u=1-x^3 and rearranging will simplify the integral to:

    -\frac13 \int \frac{1}{\sqrt[3]u} \ \mathrm{d}u = - \frac13 \int u^{-\frac13} \ \mathrm{d}u

    Solve this integral and substitute u=1-x^3 in the final answer.
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  3. #3
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    Quote Originally Posted by dankelly07 View Post
    [snip]
    integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

    This ends up in form alog3+blog2 I just need to know how to get there..
    Partial fraction decomposition is an obvious option.
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  4. #4
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    Quote Originally Posted by dankelly07 View Post
    ok I'm having problems with these if anyone can help...


    integral of x^2(1+x^3)^(-1/3)dx



    and this other one

    integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

    This ends up in form alog3+blog2 I just need to know how to get there..
    Just filling in a couple of steps...

    Find \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x

    Let u=1-x^3 so \frac{du}{dx}=-3x^2

    Rearrange your original integral so that you have \frac{du}{dx} as a multiple, in this case by multiplying by a cleverly disguised 1, namely \frac{-3}{-3}

    Now

    \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x = \int \frac{-3x^2}{-3\sqrt[3]{1-x^3}} \ \mathrm{d}x = \frac{-1}{3}\int \frac{1}{\sqrt[3]{u}} \frac{du}{dx}\ dx = \frac{-1}{3}\int \frac{1}{\sqrt[3]{u}}\ du
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