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

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SUMMARY

The integral of the function 1/(x^2 + a) can be calculated using a substitution method. By letting y = x/√a, the integral transforms into ∫ (1/(ay^2 + a))(√a dy), which simplifies to (√a/a) ∫ (1/(y^2 + 1)) dy. The result of this integral is (√a/a) arctan(y) + C, leading to the final primitive function of (1/√a) arctan(x/√a) + C for any real number a > 0.

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Holali
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Hi,
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 don't know how it is with different 'a' values.
Thanks for help!
 
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First, this should not have been posted under "differential equations". I will move it to "Calculus and Analysis".

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

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