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I know that the solution is $\frac{arctan(\frac{x}{a})}{a}$, however the thought popped into my head of trying to show this by using partial fractions and logs.

I rewrote $\frac{1}{x^2+a^2}$ as $\frac{1}{2ai}(\frac{1}{x-ai}-\frac{1}{x+ai})$. Then my chain of thought went as follows:

$\frac{1}{2ai} \int\frac{1}{x-ai}-\frac{1}{x+ai} dx$

$=\frac{1}{2ai}(log(x-ai)-log(x+ai))$

$=\frac{1}{2ai}(log(|x-ai|*e^{arg(x-ai)i})-log(|x+ai|*e^{arg(x+ai)i}))$ (rewriting the complex numbers in modulus-argument form)

$=\frac{1}{2ai}(log(e^{arg(x-ai)i})-log(e^{arg(x+ai)i}))$ (using the addition rule of logs and cancelling the resultant $log(|x-ai|)$ and $log(|x+ai|)$)

$=\frac{1}{2ai}(arg(x-ai)i-arg(x+ai)i)$

$=\frac{1}{2a}(arg(x-ai)-arg(x+ai))$

$=\frac{1}{2a}(arctan(\frac{-a}{x})-arctan(\frac{a}{x}))$

$=\frac{1}{2a}(-2arctan(\frac{a}{x}))$ (using $arctan(-x)=-arctan(x)$)

$=\frac{-1}{a}arctan(\frac{a}{x})$

Which is close to, but clearly not, the actual answer. I cannot see where I went wrong with my thinking. Can anyone tell me where my error is?

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  • $\begingroup$ General advice: do the derivative of the supposed primitive. Trigonometric things can be written in many equivalent forms. $\endgroup$ Dec 2, 2016 at 9:21

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Since $\arctan(a/x)=\frac{\pi}{2} - \arctan(x/a)$ if $x/a>0$ (and something similar if $x/a<0$), your answer actually does agree with the known one (for $x\neq 0$), up to a constant of integration.

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