Results 1 to 6 of 6

Math Help - Integral of an exponential

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    36

    Integral of an exponential

    I have been trying to evaluate:

    \int \frac{1}{\sqrt(e^x+1)} dx

    I'd appreciate it if anyone could go through this step by step? My teacher is dropping this on us the day before the test unfortunately. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Put t=\sqrt{1+e^{x}}\implies \frac{2}{t^{2}-1}\,dt=\frac{dx}{\sqrt{1+e^{x}}}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    202
    how about 1=e^x.e^-x???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2009
    Posts
    202
    after that t= e^x+1 :the substitution. u should get an integral to solve by-parts.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629
    Hello, Spudwad!

    Another approach . . .


    \int \frac{dx}{\sqrt{e^x+1}}

    Let: . u \:=\:\sqrt{e^x+1} \quad\Rightarrow\quad u^2 \:=\:e^x+1 \quad\Rightarrow\quad e^x \:=\:u^2-1 \quad\Rightarrow\quad x \:=\:\ln(u^2-1)

    . . Then: . dx \:=\:\frac{2u\,du}{u^2-1}


    Substitute: . \int\frac{\frac{2u\,du}{u^2-1}}{u} \;=\;2\int\frac{du}{u^2-1}



    At this point, you can use Partial Fractions or a standard formula

    . . and get: . \ln\left|\frac{u-1}{u+1}\right| + C


    Back-substitute: . \ln\left|\frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} + 1}\right| + C


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


    If you want to impress/surprise/terrify your teacher, simplify . . .


    Multiply the fraction by: \frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1}-1}

    . . \frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} + 1}\cdot{\color{blue}\frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} - 1}} \;=\; \frac{(\sqrt{e^x+1} - 1)^2}{(e^x+1) - 1} \;=\; \frac{(\sqrt{e^x+1} - 1)^2}{e^x}


    And the answer becomes: . \ln\left(\frac{(\sqrt{e^x+1} - 1)^2}{e^x}\right) + C

    . . . =\;\ln\left((\sqrt{e^x+1} - 1)^2\right) - \ln(e^x) + C \;=\;2\ln\left(\sqrt{e^x+1}-1\right) - x + C

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    actually, you used my substitution as well.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponential Integral
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 2nd 2011, 09:29 AM
  2. exponential integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 18th 2010, 05:44 AM
  3. exponential integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 22nd 2009, 04:09 AM
  4. exponential integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 17th 2008, 02:57 PM
  5. exponential integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 5th 2007, 07:31 PM

Search Tags


/mathhelpforum @mathhelpforum