Results 1 to 5 of 5

Math Help - Troublesome Integral 2

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    76

    Troublesome Integral 2

    Sorry, the first version was incorrect. This is the correct integral.


    <br />
\int (xe^{2x})/(2x + 1)^2 dx<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by kaiser0792 View Post
    Sorry, the first version was incorrect. This is the correct integral.


    <br />
\int (xe^{2x})/(2x + 1)^2 dx<br />
    Wolfram tells me that you need to use the Series expansion...

    integral&#x5b;&#x28;x&#x2a;Exp&#x5b;2&#x2a;x&#x5d; &#x29;&#x2f;&#x28;&#x28;2&#x2a;x &#x2b; 1&#x29;&#x5e;2&#x29;&#x5d; - Wolfram|Alpha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    76
    The thing is, I'm in a chapter in Calculus II that covers several of the integration techniques beyond the basic formulas, such as, integration by parts, trig integrals, trig substitution, partial fractions, L' Hopital's rule, etc. Sequences and Series aren't introduced until the next chapter, which is effectively Calculus III. I suppose this particular integral may have been included accidentally. However, I got close to the answer after integrating by parts several times. Suppose I could have gotten off track somewhere. I may try that technique again and see what I come up with.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Try to differentiate  \frac{ e^{2x}}{2x+1} and see what would happen !



    Consider

     \frac{xe^{2x}}{(2x+1)^2}

    Use these techniques : partial fraction + integration by parts

     = \frac{e^{2x}}{2} \left( \frac{1}{2x+1} - \frac{1}{(2x+1)^2} \right )

    the integral

     = \frac{1}{2} \int  e^{2x} \left( \frac{1}{2x+1} - \frac{1}{(2x+1)^2} \right ) ~dx

     = \frac{1}{2}  \left[ \int  e^{2x}  \frac{1}{2x+1}~dx - \int e^{2x} \frac{1}{(2x+1)^2}~dx \right ]

     = \frac{1}{2} \left[ \frac{1}{2} \frac{e^{2x}}{2x+1} + \int e^{2x} \frac{1}{(2x+1)^2}~dx -\int e^{2x} \frac{1}{(2x+1)^2}~dx \right ]

     = \frac{1}{4} \frac{e^{2x}}{2x+1}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    Posts
    76
    Thank you. I will study your solution tomorrow. This particular problem is at the end of a section dealing only with integration by parts, i.e., partial fractions come later in chapter. It may be a precursor problem to prepare for the partial fraction material. Either way, I appreciate your solution and, as I mentioned, I will study it.

    I appreciate the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Troublesome Stokes Theorem Problem.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 17th 2011, 01:31 PM
  2. [SOLVED] Matrix Spectra with Troublesome Trigonometric Term
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: June 25th 2010, 05:44 AM
  3. troublesome algebra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 30th 2009, 01:34 PM
  4. Replies: 0
    Last Post: November 26th 2009, 12:00 PM
  5. Troublesome standard deviation question
    Posted in the Statistics Forum
    Replies: 7
    Last Post: November 26th 2009, 10:14 AM

Search Tags


/mathhelpforum @mathhelpforum