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Integral with trig substitution

  1. Mar 10, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫(x+1)/((x^2+1)^2)

    2. Relevant equations



    3. The attempt at a solution

    I have been able to seperate this into 2

    ∫x/(x^2+1)^2 dx which i found to be equal to (1/2)arctanx

    and

    ∫1/(x^2+1)^2 dx which i am unable to find

    What i did was sub in x=tanθ and dx=sec^2(θ)dθ, and with this i was able to get the integral to the form

    ∫1/sec^2(θ)dθ=∫cos^2(θ)dθ =

    ∫(1+cos2θ)/2

    I then seperated this into two integrals

    ∫1/2=θ/2

    and

    ∫cos2θ/2 = sin2θ/4

    However, the answer given is (x-1)/(2(x^2+1)+C

    I am not sure how to get to this answer from

    θ/2+sin2θ/4+C, i tried subbing in θ=arctanx and i get a very messy equation

    arctanx/2+sin(2arctanx)/4+C

    I was able to simplify it to

    cosx/2sinx+sin(2cosx/sinx)/4+C, and i can't get any farther twoards the answer.

    Can anyone please explain how i get there or did i make a mistake somewhere in my work?

    Thanks very much
     
    Last edited: Mar 10, 2013
  2. jcsd
  3. Mar 10, 2013 #2

    LCKurtz

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    For what it's worth, their answer is certainly wrong because it surely must have an arctan(x) term. Maple gives:

    [(1/4)(4x-2)/(x^2+1)] + arctan(x)
     
  4. Mar 10, 2013 #3
    oops i entered the original question incorrectly, it should be 1+x on the top rather than 2+x, also the answer i wrote was for one part of the integral only, not the ansewr to the whole question, i should have clariified sorry.
     
  5. Mar 10, 2013 #4

    SammyS

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    [itex]\displaystyle \int \frac{x}{(x^2+1)^2}\,dx\ne\frac{1}{2}\arctan(x)[/itex]

    Use substitution to evaluate this integral. Let u = x2 + 1 ...


    For [itex]\displaystyle \int \frac{1}{(x^2+1)^2}\,dx\,,\ [/itex] write the integrand as [itex]\displaystyle \frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\ .\ [/itex]

    The first term gives the arctan part of the result. For the second term, do integration by parts, with u = x, dv = ...
     
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