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Integration techinque

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data
    (Given integral, and solution from Wolfram Alpha)
    integral of 4x/(5+2x+x^2)
    solution was -2ln |x^2+2x+5| - 2tan-1 (x+1/2) +c


    [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP20419ac3c2hibb2ic3200002fiag16ag9fa2563?MSPStoreType=image/gif&s=10&w=297&h=40 [Broken]


    [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP156619ac3641b9d1cd7700002f4hdf127hhag4ch?MSPStoreType=image/gif&s=63&w=255&h=53 [Broken]
    2. Relevant equations

    3. The attempt at a solution
    I took pictures of my work, please click on the link

    First problem
    Code (Text):
    http://i43.tinypic.com/33mv5l4.jpg
    ^ minor fix for the picture, instead of -4tan I got -tan because I had 1/4 outside the []
    But still, my answer is wrong. The solution showed 2ln *** -2tan ****




    Second problem

    Code (Text):
    http://i39.tinypic.com/34i3995.jpg
    I knew I could just first expand the square, and then multiple each term by x. But I want to do it this way, but how the solution was wrong?



    Third problem

    Code (Text):
    http://i44.tinypic.com/2s7g7pd.jpg
    Compare to the solution, I have an extra 3 and the sqrt(3) at the top...?


    Thank you
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 5, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I can't see your work but here's what I would do:
    [itex]x^2+ 2x+ 5= x^2+ 2x+ 1+ 4= (x+1)^2+ 4[/itex]

    If we let u= x+1, then [itex]x^2+ 2x+ 5= u^2+ 4[/itex], x= u- 1, and dx= du.

    The integral becomes
    [tex]\int\frac{4x}{x^2+ 2x+5} dx= \int \frac{4(u-1)}{u^2+ 4} du[/tex]
    [tex]= 2\int\frac{2u}{u^2+ 4}- \int \frac{1}{1 +(u/2)^2} du[/tex]

    Let v= u^2+ 4 in the first integral and do the second as an arctangent.
     
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