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  1. #1
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    How to integrate..

    How would one go about integrating this? Please provide full working.

    \int\frac{y}{(y^2+3)} dy

    My math teacher told me to let u = y^2+3 and use the derivative of that, but I still don't quite understand what to do.
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  2. #2
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    Quote Originally Posted by Exotique View Post
    How would one go about integrating this? Please provide full working.

    \int\frac{y}{(y^2+3)} dy

    My math teacher told me to let u = y^2+3 and use the derivative of that, but I still don't quite understand what to do.
    Rewrite it as

    \int{(y^2 + 3)^{-1}y\,dy}

     = \frac{1}{2}\int{(y^2 + 3)^{-1}\,2y\,dy}.


    Now let u = y^2 + 3 so that \frac{du}{dy} = 2y.

    The integral becomes

    \frac{1}{2}\int{u^{-1}\,\frac{du}{dy}\,dy}

     = \frac{1}{2}\int{u^{-1}\,du}

     = \frac{1}{2}\ln{|u|} + C

     = \frac{1}{2}\ln{|y^2 + 3|} + C

     = \frac{1}{2}\ln{(y^2 + 3)} + C since y^2 + 3 > 0 for all y.
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    Quote Originally Posted by Exotique View Post
    How would one go about integrating this? Please provide full working.

    \int\frac{y}{(y^2+3)} dy

    My math teacher told me to let u = y^2+3 and use the derivative of that, but I still don't quite understand what to do.
    You can just let u = y^2+3 straight away (without creating the derivative of the denominator in the numerator): if u = y^2+3, then \dfrac{du}{dy} = 2u \Rightarrow dy = \dfrac{du}{2y}. That means: \int\dfrac{y}{y^3+3}\;{dy}= \int\dfrac{y}{\left(u\right)2y}\;{du} = \dfrac{1}{2} \int\dfrac{y}{\left(u\right)y}\;{du} = \dfrac{1}{2} \int\dfrac{1}{\left(u\right)}\;{du} = \dfrac{1}{2}\log{u}+C \Rightarrow \int\dfrac{y}{y^3+3}\;{dy} = \dfrac{1}{2}\log\left(y^3+3\right)+C
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