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Math Help - Using integral by parts

  1. #1
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    Using integral by parts

    Hello everyone it's me again...

    For this question I have to use integral by parts to find the integral of (t+5)/(sqrt 8-t)

    I have:
    u= t+5
    v' = (8-t)^-1/2

    u'= 1
    v= 2(8-t)^1/2

    then I plugged in the numbers

    (t+5)(2(8-t))^1/2 = integral (8-t)^1/2

    I don't really know what to do could someone help me out? /:
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  2. #2
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    Re: Using integral by parts

    Integration by Parts is not the best method to use here.

    \displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}

    Let \displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*} and note that \displaystyle \begin{align*} t + 5 = 13 - u \end{align*} and the integral becomes

    \displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}

    This should now be easy to solve.
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  3. #3
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    Re: Using integral by parts

    Quote Originally Posted by Prove It View Post
    Integration by Parts is not the best method to use here.

    \displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}

    Let \displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*} and note that \displaystyle \begin{align*} t + 5 = 13 - u \end{align*} and the integral becomes

    \displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}

    This should now be easy to solve.
    Unfortunately I have to use integration by parts as instructed by my teacher but this should help me come up with the answer. Thank you.
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