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Math Help - Reduction Forumula

  1. #1
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    Reduction Forumula

    I'm trying to prove:

    \int (ln(x))^n=x(ln(x))^n-n\int (ln(x))^{n-1}dx

    I=\int(ln(x))^n=\int(ln(x))^{n-1}ln(x)dx

    It's not obvious what substition I should use here. If I let u=(ln(x))^{n-1} and dv=ln(x)dx then I have du=(n-1)(ln(x))^{n-2}\frac{1}{x}dx and v=xln(x)-x.

    This turn into a really complicated thing, and I can't see were I will be able to find I in this mess. I don't really expect anyone to prove this for me, since it's too much work. However, I just need some guidance. Is there some trick I'm not noticing?
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    I'm trying to prove:

    \int (ln(x))^n=x(ln(x))^n-n\int (ln(x))^{n-1}dx

    I=\int(ln(x))^n=\int(ln(x))^{n-1}ln(x)dx

    It's not obvious what substition I should use here. If I let u=(ln(x))^{n-1} and dv=ln(x)dx then I have du=(n-1)(ln(x))^{n-2}\frac{1}{x}dx and v=xln(x)-x.

    This turn into a really complicated thing, and I can't see were I will be able to find I in this mess. I don't really expect anyone to prove this for me, since it's too much work. However, I just need some guidance. Is there some trick I'm not noticing?
    use integration by parts with

    u=[\ln(x)]^n \implies du =\frac{n[\ln(x)]^{n-1}}{x}dx
    dv=dx \implies v=x

    I=x[\ln(x)]^n-n\int [\ln(x)]^{n-1}dx
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