Results 1 to 13 of 13

Thread: Integration by substitution involving e^2x + 4

  1. #1
    Lex
    Lex is offline
    Newbie
    Joined
    Feb 2010
    Posts
    11

    Integration by substitution involving e^2x + 4

    Hi everyone,

    I am trying to integrate the following expression:

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

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

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

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

    $\displaystyle \int$$\displaystyle \frac{1}{a}$$\displaystyle du$

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

    but I can't get any further. Any help would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Belgium
    Posts
    1,254
    Thanks
    24

    Re: Integration by substitution involving e^2x + 4

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

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

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

    Determine A and B.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: Integration by substitution involving e^2x + 4

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

    (where "u" is it should read "$\displaystyle e^{2x}$$\displaystyle +4$" but I couldn't get the LaTex maths thing to work, sorry)
    Just notice that $\displaystyle \frac{1}{e^{2x}+4}=\frac{e^{-2x}}{1+4e^{-2x}}$.
    Now it is easy.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lex
    Lex is offline
    Newbie
    Joined
    Feb 2010
    Posts
    11

    Re: Integration by substitution involving e^2x + 4

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

    If this is correct, I can't integrate the $\displaystyle \frac{1}{4u-16}$, can I?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Integration by substitution involving e^2x + 4

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

    It may be easier to write $\displaystyle \dfrac{1}{4u+16} = \dfrac{1}{4} \cdot \dfrac{1}{u+4}$
    Last edited by e^(i*pi); Aug 5th 2011 at 03:54 AM. Reason: constant of integration
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027

    Re: Integration by substitution involving e^2x + 4

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

    If this is correct, I can't integrate the $\displaystyle \frac{1}{4u-16}$, can I?
    Why not? since the denominator is linear, the substitution v= 4u- 16 is easy.
    (or write the fraction as $\displaystyle \frac{1}{4}\frac{1}{u- 4}$ and use the substitution v= u- 4.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Lex
    Lex is offline
    Newbie
    Joined
    Feb 2010
    Posts
    11

    Re: Integration by substitution involving e^2x + 4

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

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

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

    I'm a bit confused!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: Integration by substitution involving e^2x + 4

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

    If the problem is $\displaystyle \int {\frac{{dx}}{{e^{2x} + 4}}} $ then the answer is $\displaystyle \frac{1}{4}x - \frac{1}{8}\ln \left( {e^{2x} + 4} \right)+c$.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Lex
    Lex is offline
    Newbie
    Joined
    Feb 2010
    Posts
    11

    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 $\displaystyle \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C$

    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 $\displaystyle u = 1 + 4e^{-2x}$ and Plato's answer, I have no idea what's going on!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9

    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 $\displaystyle \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C$

    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 $\displaystyle u = 1 + 4e^{-2x}$ 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).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Lex
    Lex is offline
    Newbie
    Joined
    Feb 2010
    Posts
    11

    Re: Integration by substitution involving e^2x + 4

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

    Way to patronise.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9

    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).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Integration by substitution involving e^2x + 4

    An alternative:

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

    Now applying Partial Fractions

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

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

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

    So completing the integral...

    $\displaystyle \displaystyle \begin{align*}\int{\frac{2}{u(u^2 + 4)}\,du} &= \frac{1}{2}\int{\frac{1}{u}\,du} - \frac{1}{2}\int{\frac{1}{u^2 + 4}\,du} \\ &= \frac{1}{2}\ln{|u|} - \frac{1}{2}\arctan{\left(\frac{u}{2}\right)} +C \\ &= \frac{1}{2}\ln{\left|e^x\right|} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} +C \\ &= \frac{x}{2} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} + C\end{align*}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. First order ODE involving substitution
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Jul 21st 2011, 11:54 PM
  2. Second order D.E involving substitution
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Mar 25th 2011, 01:24 PM
  3. Integral involving trig substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 27th 2009, 04:36 PM
  4. Integration (substitution)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 31st 2009, 08:58 PM
  5. Replies: 1
    Last Post: Dec 2nd 2008, 02:51 PM

Search tags for this page


/mathhelpforum @mathhelpforum