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How to integrate function 1/(x^2+a) ?

  1. Sep 30, 2011 #1
    I would like to calculate one integral. I just want to get primitive function, not definite integral.

    ∫ 1/(x^2+a) dx

    where a is real number and >0

    I only found that

    ∫ 1/(x^2+1) dx ,
    its arctan(x) + C,
    but i dont know how it is with different 'a' values.
    Thanks for help!
  2. jcsd
  3. Sep 30, 2011 #2


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    Science Advisor

    First, this should not have been posted under "differential equations". I will move it to "Calculus and Analysis".

    Second, let [itex]y= x/\sqrt{a}[/itex] so that [itex]x= \sqrt{a}y[/itex] and [itex]dx= \sqrt{a}dy[/itex]. The integral becomes
    [tex]\int\frac{1}{ay^2+ a}(\sqrt{a}dy)= \frac{\sqrt{a}}{a}\int \frac{1}{y^2+1}dy[/tex].
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