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By partial fractions I've got $\frac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$ Then integrating $\frac{1}{2i}\int\left(\frac{1}{x-i}-\frac{1}{x+i}\right)dx=\frac{1}{2i}(\ln(x-i)-\ln(x+i))$ If $x-i=R_1e^{i\phi_1}$ and $x+i=R_2e^{i\phi_2}$ then $\frac{1}{2i}(\ln(x-i)-\ln(x+i))=\frac{1}{2i}(\ln(R_1)-\ln(R_2)+i(\phi_1-\phi_2))=\frac{1}{2}(\phi_1-\phi_2)$ But now I put that $\phi_1=\arctan\frac{-1}{x}$ and $\phi_2=\arctan\frac{1}{x}$ so: $\arctan(x)=^?\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)$ I tried some values and looks false. The most confusing thing is that $\frac{d}{dx}\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)=\frac{1}{1+x^2}$(Something is happening with the argument function.) What's the mistake?

EDIT: Hans Lundmark wisely said in a comment there's only thing left: the constant $\arctan(x)=\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)+\frac{\pi}{2}$

Now the question is: how should I find this constant in future integrations?

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    Looks like the constant is $\frac{\pi}{2}$. Thanks.2012-08-27

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Your error is in the second step: $\frac{1}{2i}\int(\frac{1}{x-i}-\frac{1}{x+i})dx=\frac{1}{2i}(\ln(x-i)-\ln(x+i))\color{red}{+C}$

When you do an indefinite integral, never forget the $+C$ part. $C$ here is an arbitrary constant. Also if you do the indefinite integral directly by knowing $\arctan'(x) = 1/(1+x^2)$ you still get $\int 1/(1+x^2)\,\mathrm dx = \arctan(x)+C$.

As how to find the constant: You don't. There's not a single constant which is the correct one for the indefinite integral. All constants are allowed (thus the $+C$). If you have a definite integral, you have $\int_a^b f(x)\,\mathrm dx = (F(b)+C)-(F(a)+C) = F(b)-F(a)$, so the constant simply cancels out.

Of course if you want a certain additional condition (such as, $F(0)=0$), then you can use that to fix the constant. But that's then due to an additional constraint of the solution, not due to the integral itself.

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    @dotdot: I was adding text about that during the time you posted the comment. The short answer is: There *is* no "correct" value of the constant to discover.2012-08-27