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Thread: Evaluating an integral

  1. #1
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    Evaluating an integral

    Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways.

    a) Integration by parts
    b) The substitution u = sqrt(x^2+1)

    Not sure how to do this, help would be greatly appreciated!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by brumby_3 View Post
    Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways.

    a) Integration by parts
    b) The substitution u = sqrt(x^2+1)

    Not sure how to do this, help would be greatly appreciated!
    a) by parts...

    ... taking into account the identity...

    (1)

    ... is...



    (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Quote Originally Posted by brumby_3 View Post

    b) The substitution u = sqrt(x^2+1)
    $\displaystyle \begin{aligned}& u: = \sqrt{x^2+1} \Rightarrow \frac{du}{dx} = \frac{2x}{2\sqrt{x^2+1}} = \frac{x}{\sqrt{x^2+1}} \Rightarrow dx = \frac{\sqrt{x^2+1}}{x}\;{du} \\& \therefore \int\frac{x^3}{\sqrt{x^2+1}}\;{dx} = \int\frac{x^3\sqrt{x^2+1}}{x\sqrt{x^2+1}}\;{du} = \int x^2\;{du} = \int (u^2-1)\;{du} = \cdots \end{aligned}$
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