1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook