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Integration by Parts help

  1. Jan 25, 2012 #1
    Hello. I'm attempting to integrat ∫ln(x+x^2)dx
    Our professor gave us the hint of x(1+x)
    I believe u= ln(x+x^2) and du=1+2x/x+x^2
    I am not sure what dv should be
    Any help would be greatly appreciated! Thanks
     
  2. jcsd
  3. Jan 25, 2012 #2

    epenguin

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    I would break your expression up a bit first into simpler things. What is the log of a product?

    Though there may be more than one way.
     
  4. Jan 25, 2012 #3
    i think this answer is:
    1/4x(-x+2(x+2)lnx-4)+C
     
  5. Jan 25, 2012 #4
    u=ln(x^2 + 1)-----)du=2x/(x^2+1)
    v=x--------------)dv= dx.

    note: der( u*v) = u dv+ v du.
    and then Integrate both sides and you get :
    uv= int(u dv)+ int (v du). Switch it around and you get int(u dv) =uv- int (v du)
    So the integral is ln(x^2 + 1)*x- int(2x^2/x^2+1)
    Next, integrate the last part. int(2x^2/(x^2+1)). Take the 2 out for a minute. Then add a one and subtract a one in the top. int(x^2+1-1)/(x^2+1). Separate them and the integral is int(1-1/(x^2+1)). That's x - tan inverse x. Put the 2 back in, and it's 2x-2 taninvers x.
    Put them together and you will have ln(x^2 + 1)*x-2x+2 taninverse x+C.
     
  6. Jan 25, 2012 #5

    HallsofIvy

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    Ouch! Where did [itex]ln(x^2+ 1)[/itex] come from? That is much harder than the given [itex]ln(x^2+ x)[/itex]

    What epenguin was suggesting was that you use the fact that [itex]ln(x+ x^2)= ln(x(1+ x))= ln(x)+ ln(1+ x)[/itex].
     
    Last edited by a moderator: Jan 25, 2012
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