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Thread: Reduction formula

  1. #1
    Junior Member symstar's Avatar
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    Reduction formula

    I need to use the reduction formula for this problem.
    $\displaystyle \int(\ln{x})^{n} dx$ where $\displaystyle n\geqslant{2}$

    Would these be the best choices for the integration?
    $\displaystyle u=(\ln{x})^{n-1}$
    $\displaystyle du=(\tfrac{1}{x})(n-1)(\ln{x})^{n-2} dx$
    $\displaystyle dv=\ln{x} dx$
    $\displaystyle v=x\ln{x}-x$
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  2. #2
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    Quote Originally Posted by symstar View Post
    I need to use the reduction formula for this problem.
    $\displaystyle \int(\ln{x})^{n} dx$ where $\displaystyle n\geqslant{2}$

    Would these be the best choices for the integration?
    $\displaystyle u=(\ln{x})^{n-1}$
    $\displaystyle du=(\tfrac{1}{x})(n-1)(\ln{x})^{n-2} dx$
    $\displaystyle dv=\ln{x} dx$
    $\displaystyle v=x\ln{x}-x$
    No.

    Let $\displaystyle I_n = \int (\ln x)^{n} \, dx$.

    Use the choice $\displaystyle u = (\ln x)^{n}$ and $\displaystyle dv = dx$.

    Then $\displaystyle I_n = x \, (\ln x)^n - n I_{n-1} \, ....$
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