Why can't I solve the integral $\int (1+x^2)^{-1}dx$ this way? Or at least, I've been told that this way is wrong, why is that?
$$\int (1+x^2)^{-1}dx=\frac i 2\int (x+i)^{-1}dx-\frac i 2 \int (x-i)^{-1}dx=\frac i 2 \left(\ln\left(\frac {x+i}{x-i}\right)\right)$$
You can, just make sure you have a $+C$ because logs and complex numbers are ugly wrt branch cuts.
What you wrote is another form of $\tan^{-1}(x)$. In general though, logs and complex numbers shouldn't be mixed unless you know exactly what you're doing and are careful with branch cuts.
You can determine an antiderivative that way; however, you must take care when dealing with logarithms of complex numbers as the argument isn't well-defined.
What you have found is just an alternative form of $\arctan x$.
