Path independence of an integral? I'm studying for a test (that's why I've been asking so much today,) and one of the questions is about saying if an integral is path independent and then solving for it. 
I was reading online about path independence and it's all about vector fields, and I'm very, very lost.
This is the integral
$$\int_{0}^{i} \frac{dz}{1-z^2}$$
So should I find another equation that gives the same result with those boundaries? I honestly just don't know how to approach the problem, any links or topics to read on would be appreciated as well.
Thank you!
 A: You can simply do the following. Consider two paths joining $0$ and $i$: one will be a vertical segment, and another one going somehow around the point $z=1$ (for example, a counterclockwise sequence of three segments $0\rightarrow 1-i\rightarrow 2+i\rightarrow i$. The difference of integrals along these two paths will be given by an integral along a closed contour around $z=1$, which can be evaluated by Cauchy integral formula to be
$$ \oint_{z=1}\frac{dz}{1-z^2}=2\pi i \left[\frac{1}{z+1}\right]_{z=1}=\pi i.$$
So the integral is not path independent.
In general, if the integrand has at least one simple pole, the integral will depend on the chosen path.
A: Seeing other answers, the follםwing perhaps doesn't grab the OP's intention, but here it is anyway.
Putting $\,z=x+iy\implies\,z^2=x^2-y^2+2xyi\,$ , so along the $\,y$-axis from zero to $\,i\,$ we get:
$$x=0\;,\;\;0\le y\le 1\implies \frac1{1-z^2}=\frac1{1+y^2}\;,\;\;dx=0 \;,\;\;dz=i\,dy\;,\;\ \;\text{so}$$
$$\int\limits_0^i\frac{dz}{1-z^2}=\left.\int\limits_0^1\frac{i\,dy}{1+y^2}= i\arctan y\right|_0^1=\frac\pi 4i$$
A: There's an easy analytic primitive of $\frac 1 2 \log (1+z) - \frac 1 2 \log (1-z)$ on $\mathbb C \setminus \mathbb R \cup (-1,1)$ if you pick the correct branch of the logarithm. So it's certainly path independent if the path doesn't pass through the wrong parts of $\mathbb R$, i.e. don't pass through $|r|\geq 1$
