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Math Help - Integration by partial fractions help?

  1. #1
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    Integration by partial fractions help?

    I'm trying to find the integral of

    (-12e^x-20)/(e^(2x)+6e^x+5)

    using integration by partial fractions, but I keep getting stuck at

    -4 (/int/ (1/2)/(e^x+1) + /int/ (5/2)/(e^x+5))

    I'm not sure if I messed up on one of the intermediate steps. If not, how do I proceed from here to finish integrating?

    Note: the answer is supposed to be -4x+2ln(1+e^x)+2ln(5+e^x), but I have no idea how to get there.

    Any help would be appreciated, thank you!
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  2. #2
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    Re: Integration by partial fractions help?

    Quote Originally Posted by jiel View Post
    I'm trying to find the integral of

    (-12e^x-20)/(e^(2x)+6e^x+5)

    using integration by partial fractions, but I keep getting stuck at

    -4 (/int/ (1/2)/(e^x+1) + /int/ (5/2)/(e^x+5))

    I'm not sure if I messed up on one of the intermediate steps. If not, how do I proceed from here to finish integrating?

    Note: the answer is supposed to be -4x+2ln(1+e^x)+2ln(5+e^x), but I have no idea how to get there.

    Any help would be appreciated, thank you!
    $f(x)=\dfrac{-12 e^x - 20}{e^{2x}+6e^x+5}$

    let $u=e^x$

    $f(u)=\dfrac{-12u-20}{u^2+6u+5}$

    $f(u)=\dfrac{-12u-20}{(u+5)(u+1)}$

    $f(u) = \dfrac A {u+5} + \dfrac B {u+1}$

    $A(u+1) + B(u+5) = -12u - 20$

    You should be able to finish this from here. Just substitute back $e^x$ for $u$ when you are finished.
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  3. #3
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    Re: Integration by partial fractions help?

    Hello, jiel!

    I'm trying to find the integral of: \frac{-12e^x-20}{e^{2x}+6e^x+5} with partial fractions

    but I keep getting stuck at: . -4\int\left(\frac{\frac{1}{2}}{e^x+1} +\frac{\frac{5}{2}}{e^x+5}\right)dx
    Your work is correct!

    You have: . -2\int\left(\frac{1}{e^x+1} + \frac{5}{e^x+5}\right)dx


    Watch this!

    Consider: . I \;=\;\int\frac{dx}{e^x+1}

    Multiply by \frac{e^{-x}}{e^{-x}}\!:\;\;I \;=\;\int\frac{e^{-x}}{e^{-x}}\cdot\frac{dx}{e^x+1} \;=\;\int\frac{e^{-x}dx}{1 + e^{-x}}

    Let u \,=\, 1+e^{-x} \quad\Rightarrow\quad du \,=\,-e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \,=\,-du

    Substitute: . I \;=\;\int\frac{-du}{1+u} \;=\;-\int\frac{du}{1+u} \;=\;-\ln|1+u| + C

    Back-substitute: . I \;=\;-\ln\left(1 + e^{-x}\right) + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This can be simplified beyond all recognition . . .

    -\ln(1 + e^{-x}) + C

    . . . =\;-\ln\left(1 + \frac{1}{e^x}\right) + C

    . . . =\;-\ln\left(\frac{e^x+1}{e^x}\right) + C

    . . . =\;\ln\left(\frac{e^x+1}{e^x}\right)^{-1} + C

    . . . =\; \ln\left(\frac{e^x}{e^x+1}\right)+C

    . . . =\;\ln(e^x) - \ln(e^x+1) + C

    . . . =\;x - \ln(e^x+1) + C
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