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Math Help - Integration by substitution involving e^2x + 4

  1. #1
    Lex
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    Integration by substitution involving e^2x + 4

    Hi everyone,

    I am trying to integrate the following expression:

    \int \frac{2}{u} \mathrm{d}x

    (where "u" is it should read " e^{2x} +4" but I couldn't get the LaTex maths thing to work, sorry)

    using the given substitution u = e^{2x} +4

    Please can anyone offer a way to do this? I get to the stage where I have:

    \int \frac{1}{a} du

    (where "a" is it should read "u(u-4)")

    but I can't get any further. Any help would be much appreciated.
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    Re: Integration by substitution involving e^2x + 4

    If you have to integrate:
    \int \frac{2dx}{e^{2x}+4}

    If you say, let u=e^{2x}+4 \Rightarrow du=2\cdot e^{2x}dx \Leftrightarrow \frac{du}{u-4}=2dx, you'll get:
    \int \frac{2dx}{e^{2x}+4}=\int \frac{du}{u\cdot (u-4)}

    Try to split up in partial fractions:
    \frac{1}{u\cdot(u-4)}=\frac{A}{u}+\frac{B}{u-4}

    Determine A and B.
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Lex View Post
    \int \frac{2}{u} \mathrm{d}x

    (where "u" is it should read " e^{2x} +4" but I couldn't get the LaTex maths thing to work, sorry)
    Just notice that \frac{1}{e^{2x}+4}=\frac{e^{-2x}}{1+4e^{-2x}}.
    Now it is easy.
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  4. #4
    Lex
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    Re: Integration by substitution involving e^2x + 4

    Siron, I got \frac{1}{u\cdot(u-4)}=\frac{1}{4u-16}-\frac{1}{4u}

    If this is correct, I can't integrate the \frac{1}{4u-16}, can I?
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    Re: Integration by substitution involving e^2x + 4

    Sure you can: \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C

    It may be easier to write \dfrac{1}{4u+16} = \dfrac{1}{4} \cdot \dfrac{1}{u+4}
    Last edited by e^(i*pi); August 5th 2011 at 04:54 AM. Reason: constant of integration
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Lex View Post
    I got \frac{1}{u\cdot(u-4)}=\frac{1}{4u-16}-\frac{1}{4u}

    If this is correct, I can't integrate the \frac{1}{4u-16}, can I?
    Why not? since the denominator is linear, the substitution v= 4u- 16 is easy.
    (or write the fraction as \frac{1}{4}\frac{1}{u- 4} and use the substitution v= u- 4.)
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  7. #7
    Lex
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Plato View Post
    Just notice that \frac{1}{e^{2x}+4}=\frac{e^{-2x}}{1+4e^{-2x}}.
    Now it is easy.
    I used the substitution u = 1 + 4e^{-2x} and got:

    -\frac{1}{8}ln(e^{2x}+4) + c

    My book says the answer is \frac{1}{2}x - \frac{1}{4}ln(e^{2x}+4) + c

    I'm a bit confused!
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Lex View Post
    I used the substitution u = 1 + 4e^{-2x} and got: -\frac{1}{8}ln(e^{2x}+4) + c
    My book says the answer is \frac{1}{2}x - \frac{1}{4}ln(e^{2x}+4) + c I'm a bit confused!
    I second that.

    If the problem is \int {\frac{{dx}}{{e^{2x}  + 4}}} then the answer is \frac{1}{4}x - \frac{1}{8}\ln \left( {e^{2x}  + 4} \right)+c.
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  9. #9
    Lex
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    Re: Integration by substitution involving e^2x + 4

    OK...

    I get the right answer (or at least the same answer as the book) when I use \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C

    I also get the right answer when I use the substitution v = 4u - 16.

    So thanks to e^(i*pi) and HallsofIvy.

    As for using the substitution u = 1 + 4e^{-2x} and Plato's answer, I have no idea what's going on!!
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Lex View Post
    OK...

    I get the right answer (or at least the same answer as the book) when I use \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C

    I also get the right answer when I use the substitution v = 4u - 16.

    So thanks to e^(i*pi) and HallsofIvy.

    As for using the substitution u = 1 + 4e^{-2x} and Plato's answer, I have no idea what's going on!!
    Read Plato's post again. Look at the numerator of the integrand. So multiply his answer by 2 (I think Plato misread the question - an easy mistake to make given the confusing way the questin was posted).
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  11. #11
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    Re: Integration by substitution involving e^2x + 4

    OK, yes, I see that the 2 was missing.

    Way to patronise.
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    Re: Integration by substitution involving e^2x + 4

    Quote Originally Posted by Lex View Post
    OK, yes, I see that the 2 was missing.

    Way to patronise.
    Excuse me??

    Multiply the numerator and denominator of your integrand by e^(-2x) to get what Plato has. Then make the substitution u = 1 + 4e^(-2x) and the answer is immediate. ln|1 + 4 e^(-2x)| and ln|e^(2x) + 4)|-2x are equivalent by the way ...... (Use basic log rules to prove this).
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    Re: Integration by substitution involving e^2x + 4

    An alternative:

    \displaystyle \begin{align*} \int{\frac{2}{e^{2x} + 4}\,dx} &= \int{\frac{2e^x}{e^{3x} + 4e^x}\,dx} \\ &= \int{\frac{2}{u^3 + 4u}\,du}\textrm{ after making the substitution }u = e^x \implies du = e^x\,dx \\ &= \int{\frac{2}{u(u^2 + 4)}\,du} \end{align*}

    Now applying Partial Fractions

    \displaystyle \begin{align*} \frac{A}{u} + \frac{Bu + C}{u^2 + 4} &= \frac{2}{u(u^2 + 4)} \\ \frac{A(u^2 + 4) + (Bu + C)u}{u(u^2 + 4)} &= \frac{2}{u(u^2 + 4)} \\ A(u^2 + 4) + (Bu + C)u &= 2 \\ Au^2 + 4A + Bu^2 + Cu &= 2 \\ (A + B)u^2 + Cu + 4A &= 0u^2 + 0u + 2 \end{align*}

    So \displaystyle 4A = 2 \implies A = \frac{1}{2}, C = 0, A + B = 0 \implies \frac{1}{2} + B = 0 \implies B = -\frac{1}{2}.

    Therefore \displaystyle \frac{\frac{1}{2}}{u} + \frac{-\frac{1}{2}u}{u^2 + 4} = \frac{2}{u(u^2 + 4)}

    So completing the integral...

    \displaystyle \begin{align*}\int{\frac{2}{u(u^2 + 4)}\,du} &= \frac{1}{2}\int{\frac{1}{u}\,du} - \frac{1}{2}\int{\frac{1}{u^2 + 4}\,du} \\ &= \frac{1}{2}\ln{|u|} - \frac{1}{2}\arctan{\left(\frac{u}{2}\right)} +C \\ &= \frac{1}{2}\ln{\left|e^x\right|} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} +C \\ &= \frac{x}{2} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} + C\end{align*}
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