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Math Help - integral of( 8x+10)/(sqrt(5-2x-x^2)

  1. #1
    Boo
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    integral of( 8x+10)/(sqrt(5-2x-x^2)

    Please, if I have
    \int \frac{8x-10}{\sqrt{5-2x-x^2}}

    how do I get rid of the sqrt in denominator?
    Thanks!!
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    Re: integral of( 8x+10)/(sqrt(5-2x-x^2)

    integral of( 8x+10)/(sqrt(5-2x-x^2)-integral-3.png
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    Re: integral of( 8x+10)/(sqrt(5-2x-x^2)

    Quote Originally Posted by Boo View Post
    Please, if I have
    \int \frac{8x-10}{\sqrt{5-2x-x^2}}

    how do I get rid of the sqrt in denominator?
    Thanks!!
    \displaystyle \begin{align*} \int{ \frac{8x - 10}{\sqrt{5 - 2x - x^2}} \,dx } &= \int{\frac{8x + 8}{\sqrt{5 - 2x - x^2}}\,dx} -  \int{\frac{18}{\sqrt{-\left( x^2 + 2x - 5 \right)}}\,dx} \\ &= -4\int{\frac{-2x - 2}{\sqrt{5 - 2x - x^2}}} - \int{\frac{18}{\sqrt{-\left( x^2 + 2x + 1^2 - 1^2 - 5 \right)}}\,dx} \\ &= -4\int{ \frac{-2x-2}{\sqrt{5 - 2x - x^2}} \,dx } - \int{\frac{18}{\sqrt{-\left[ \left( x + 1 \right)^2 - 6 \right]}} \,dx} \\ &= -4\int{\frac{-2x-2}{\sqrt{5-2x-x^2}}\,dx} - \int{\frac{18}{\sqrt{6 - \left( x + 1 \right)^2}}\,dx} \end{align*}

    In the first integral, make the substitution \displaystyle \begin{align*} u = 5-2x-x^2 \implies du = -2x-2\,dx \end{align*} and in the second, make the substitution \displaystyle \begin{align*} x + 1 = \sqrt{6}\sin{\theta} \implies dx = \sqrt{6}\cos{\theta}\,d\theta \end{align*}. See how you go
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