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Thread: Integration problem

  1. #1
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    Integration problem

    I've worked through a long substitution question and have come to this point.

    t^2(du/dt) = 2te^(-u)

    I know I have to integrate both sides but am struggling with how to do it?!
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  2. #2
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    $\displaystyle \displaystyle t^2\frac{du}{dt} = 2te^{-u}$

    Let's start by cancelling t from both sides,

    $\displaystyle \displaystyle t\frac{du}{dt} = 2e^{-u}$

    Now we'll isolate the derivative as this may be separable.

    $\displaystyle \displaystyle \frac{du}{dt} = \frac{2e^{-u}}{t}$

    Seems we can now separate so,

    $\displaystyle \displaystyle \frac{du}{2e^{-u}} = \frac{dt}{t}$

    Neatening up a little

    $\displaystyle \displaystyle \frac{1}{2}e^{u}~du = \frac{dt}{t}$

    And integrating

    $\displaystyle \displaystyle \int \frac{1}{2}e^{u}~du = \int \frac{dt}{t}$

    $\displaystyle \displaystyle \frac{1}{2}e^{u} = \ln (t)+C$

    Finishing,

    $\displaystyle \displaystyle e^{u} = 2\ln (t)+C$

    $\displaystyle \displaystyle u = \ln{( 2\ln (t)+C)}$
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