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Thread: Integration using partial fractions

  1. #1
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    Integration using partial fractions

    S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.

    Thanks!
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  2. #2
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    Quote Originally Posted by JYZero View Post
    S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.

    Thanks!
    Let $\displaystyle \frac{A}{t + 4} + \frac{B}{t - 1} = \frac{1}{(t + 4)(t - 1)}$

    $\displaystyle \frac{A(t - 1) + B(t + 4)}{(t + 4)(t - 1)} = \frac{1}{(t + 4)(t - 1)}$

    $\displaystyle A(t - 1) + B(t + 4) = 1$

    $\displaystyle At - A + Bt + 4B = 1$

    $\displaystyle (A + B)t + 4B - A = 0t + 1$.


    So $\displaystyle A + B = 0$ and $\displaystyle 4B - A = 1$


    $\displaystyle 4B - 1 = A$


    So $\displaystyle 4B - 1 + B = 0$

    $\displaystyle 5B - 1 = 0$

    $\displaystyle B = \frac{1}{5}$



    $\displaystyle A + B = 0$

    $\displaystyle A + \frac{1}{5} = 0$

    $\displaystyle A = -\frac{1}{5}$.



    So $\displaystyle \int{\frac{1}{(t + 4)(t - 1)}\,dt} = \int{-\frac{1}{5}\left(\frac{1}{t + 4}\right) + \frac{1}{5}\left(\frac{1}{t - 1}\right)\,dt}$

    $\displaystyle = \frac{1}{5}\int{\frac{1}{t - 1} - \frac{1}{t + 4}\,dt}$

    $\displaystyle = \frac{1}{5}\left[\ln{|t - 1|} - \ln{|t + 4|}\right] + C$

    $\displaystyle = \frac{1}{5}\ln{\left|\frac{t - 1}{t + 4}\right|} + C$.
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  3. #3
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    Quote Originally Posted by JYZero View Post
    S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.
    $\displaystyle \frac{1}{(t+4)(t-1)}=\frac{1}{5}\cdot \frac{(t+4)-(t-1)}{(t+4)(t-1)}=\frac{1}{5}\left( \frac{1}{t-1}-\frac{1}{t+4} \right).$
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