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Math Help - Integrals

  1. #1
    TGS
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    Integrals

    I just want to know if my final answer is correct since I don't have the solution for it.

    \int\frac{\sqrt{x} - 2x^2}{x} dx

    I obtained:  \frac{2x^\frac{3}{2}}{3}lnx - \frac{2x^3}{3} + C

    Also, I need some help figuring out this integral:

    \int x(x^2 + 1)^5 dx It is continuous from 0 to 1.
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  2. #2
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    Quote Originally Posted by TGS View Post
    I just want to know if my final answer is correct since I don't have the solution for it.

    \int\frac{\sqrt{x} - 2x^2}{x} dx

    I obtained:  \frac{2x^\frac{3}{2}}{3}lnx - \frac{2x^3}{3} + C

    Also, I need some help figuring out this integral:

    \int x(x^2 + 1)^5 dx It is continuous from 0 to 1.
    No, the first is not correct.

    Divide everything by x to turn the integral into

    \int{x^{-\frac{1}{2}} - 2x\,dx}

     = 2x^{\frac{1}{2}} - x^2 + C

     = 2\sqrt{x} - x^2 + C.


    For the second, you need to use a u substitution.

    Let u = x^2 + 1 so that \frac{du}{dx} = 2x.

    So your integral becomes

    \int{x(x^2 + 1)^5\,dx} = \frac{1}{2}\int{2x(x^2 + 1)^5\,dx}

     = \frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}

     = \frac{1}{2}\int{u^5\,du}

     = \frac{1}{12}u^6 + C

     = \frac{1}{2}(x^2 + 1)^6 + C.


    Alternatively, you could expand everything, but the expansion of (x^2 + 1)^5 is difficult unless you know the Binomial Theorem.
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