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Math Help - Easy Indefinite Integral

  1. #1
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    Easy Indefinite Integral

    Hello,
    how do I find the indefinite integral? What do I set to be equal to u?

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  2. #2
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    Re: Easy Indefinite Integral

    You shouldn't use substitution here.

    \displaystyle \begin{align*} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= \int{\frac{x^2 + 5x - 8}{x^{\frac{1}{2}}}\,dx} \\ &= \int{x^{2 - \frac{1}{2}} + 5x^{1 - \frac{1}{2}} - 8x^{-\frac{1}{2}} \,dx} \\ &= \int{x^{\frac{3}{2}} + 5x^{\frac{1}{2}} - 8x^{-\frac{1}{2}}\,dx} \end{align*}

    You can now use the power rule on each term.

    If you REALLY want to use substitution, rewrite the integral as \displaystyle \begin{align*} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= 2\int{\frac{\left( \sqrt{x} \right) ^4 + 5 \left( \sqrt{x} \right) ^2 - 8}{2\sqrt{x}}\,dx} \end{align*} and then let \displaystyle u = \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx and the integral becomes \displaystyle 2\int{u^4 + 5u^2 - 8 \, du}.

    I'm sure whichever method you choose you can now finish the problem off.
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  3. #3
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    Re: Easy Indefinite Integral

    complete the square then use trig substitution

    oh ^ works too
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