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Help with integration

  1. Jun 27, 2006 #1

    I have a question regarding some simple integrals. For the life of my I cannot seem to get the right answer. I know the solution to the problem, but my answer seems to missing an ever so important (1/2) from the equation. Here is what I have:

    The probelm is this, ∫tan^(-1)x dx.
    [This is the indefinate integral of the inverse tangent of x]

    When I work the problem out, my answer comes out to be:

    xtan^(-1)x - ln|1+x²|+C

    However, the answer that my professor and calculator give me is

    xtan^(-1)x - (1/2)ln|1+x²|+C

    Any thoughts of where this (1/2) is coming from?

    To come to this answer, I first used integration by parts (uv-∫vdu) and used tan^(-1)x as u and (1/(1+x²)) for du. I then took x^0 for dv and x as v.

    The next substitution I made was to set x = tan(θ) which made dx = sec²(θ) dθ.

    To find θ after I completed the integration, I made a triangle and found that sec(θ) was 1+x². Substituting that back into the equation gives me my final answer of:

    xtan^(-1)x - ln|1+x²|+C.

    I hope that explaination made sense. I can explain in more depth if needed.

    Thank you,

  2. jcsd
  3. Jun 27, 2006 #2


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    You don't have to do nearly so much work in the second integral--just substitute u = x^2 + 1. Most likely your error was in forgetting to introduce a square root sign when you determine sec(θ), but I'm not sure where that comes in at all since when I do it with your substitution I have to determine cos(θ) instead.
    Last edited: Jun 27, 2006
  4. Jun 27, 2006 #3

    Thanks for the reply. I tried to substitute u = x^2 + 1, but then du will then = 2x dx no? The second integral does not have another x to use. I don't know if I am saying that clearly or not. I guess what I am trying to say is that if the integral is to evaluate 1/(1+x^2) dx, then you cannot choose to substitute u as 1 + x^2 because du will not make sence. That is why you have to solve the eqation with trig substitution.

    If I am totally off here, please let me know. This is just my thinking and if you know a way so that I do not have to use trig sub and just u sub then that would be awesome.

    As for the second part of your response, I don't see where I should have a square root sign. Is there any way I could ask you to elaborate a little more?

    Thank you so much for your help so far, I really appreciate everything.

  5. Jun 27, 2006 #4


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    Dearly Missed

    No, Prace.
    You have clearly misunderstood how to do this, so I'll go through the details:
    The substitution [itex]u=1+x^{2}\to{du}=2xdx[/itex] leads to:
  6. Jun 27, 2006 #5
    You definitely do NOT have to usea trig substitution.

    If u = 1 + x2

    du = 2x*dx
    so it follows that
    du/2 = x*dx which is what you should have in that integral and so you can substitute du/2 in.
  7. Jun 27, 2006 #6
    OH MAN!! Wow, I can't believe I overlooked that. It completely makes sense now. Arrr... I hate errors like that where it is not misunderstanding of the technique or the problem, but a careless error that could have totally been avoided. Thank you so much for everyone's help and sorry if I wasted peoples time with careless errors.
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