Integration with complex constants So I saw a youtube video of the following integral, which can be solved by letting u=tanx, then you get
$\int \frac{1}{1+x^2} dx $ = ${tan}^{-1}x + C$
But it should also be solvable through partial fraction decomposition.
$\int \frac{1}{1+x^2}dx$ = $\int {\frac{A}{x-i} + \frac{B}{x+i}dx}$
$\frac{1}{1+x^2}$ = $\frac{A}{x-i}$ + $\frac{B}{x+i}$
$1$ = $A(x+i)$ + $B(x-i)$
$x=i$:  $1$ = $2iA$
$A$=$\frac 1{2i}$ = -$\frac {i}{2}$
$x=-i$: $1 = -2iB$
$B$ = $\frac{i}2$
$\int \frac{1}{1+x^2}dx$ = $\int -\frac{i}{2(x-i)}$ + $ \frac{i}{2(x+i)}dx$ = $-\frac{i}2$ln$\lvert {x-i}\rvert$ + $\frac{i}2$$ln\lvert {x+i}\rvert$ + C = $\frac{i}2ln\lvert {x+i}\rvert - ln\lvert {x-i}\rvert$ + C
This should be equal to ${tan}^{-1}x + C$ right?
${tan}^{-1}({1})$ =  $\frac\pi4$
$\frac{i}2(ln\lvert {1+i}\rvert - ln\lvert {1-i}\rvert)$ = $\frac{i}2(ln{\sqrt2}-ln{\sqrt2})$=O
So these are obviously not equal, then I thought why must i have absolute value sign in the logarithmic expressions, that rule I know only applies for integarting with real constants.
Thus,
$\int \frac{1}{1+x^2} dx = $$\frac{i}2(ln({x+i}) - ln({x-i}))$ + C = $\frac{i}2ln(\frac{({x+i})^2}{x^2+1}) + C$
Evaluated at x=1 gives us
$\frac{i}2ln(\frac{({1+i})^2}{1^2+1})$ = $\frac{i}2ln(i)$ = $\frac{i}2\cdot \frac{i\pi}{2}$ = $-\frac\pi4$
So, im quite close, but I must have missed something somewhere, and I cant find the mistake. But this triggers two other questions: How would I show that
$\frac{i}2ln(\frac{({x+i})^2}{x^2+1})$ = $\tan^{-1}x$ (or some other logarithmic expression if that expression is where the error lies)
and, why dont I need to take absolute value when integrating
$\int \frac{1}{x+a}dx$
with complex constants? Is the explanation as easy that I can evaluate all logarithmic expressions using complex numbers?

$$e^{xi} = \cos{x}+i\sin{x}$$
$$e^{-xi} = \cos{x}-i\sin{x}$$
$$\sin{x} =  \frac{ e^{xi}-e^{-xi} }{2i}$$
$$ \cos{x} = \frac{ e^{xi}+e^{-xi} }{2}$$
$$\tan{x} = \frac{\sin{x}}{\cos{x}} =\frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
$$\tan{x} = \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
Say $ \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$
$$ \tan{x} = y , x = \arctan{y}$$
$$ \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$$
Say $x = i \log{z}$
$$ \frac{1}{i}  \frac{ z^{-1}-z }{z^{-1}+z} = y$$
$$ z^{-1}-z = i y (z^{-1}+z) $$
$$1-z^2 = i y (1+z^2)$$
$$i y z^2+z^2 = 1-i y$$
$$ z = \sqrt{ \frac{ 1-i y}{1+i y} }$$
$$\arctan{y} = i \log{  \sqrt{ \frac{ 1-i y}{1+i y} }}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ 1-i y}{1+i y }}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ (1-i y)^2}{(1+i y )(1-i y)}}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ 1-2y i -y^2}{1+y^2 }}$$
