i know how it solved with re^iα but i have to solve it in cartesian form;
$z=x+iy$
\[\int {dz\over z}= \int {x dx+y dy+i(xdy-ydx) \over x^2+y^2}= i \int xdy-ydx= -i \int {dx \over y} -i \int_1^1 {dx \over \sqrt {1-x^2}} \]
(i know the final answer is $2\pi i$) I need help for rest of solution
Since $xdy=\frac{x}{y}\underbrace{ydy}_{-xdx}=-\frac{x^2}{y}dx$,$$\oint\frac{dz}{z}=-i\int\frac{y+x^2/y}{x^2+y^2}dx=-i\int\frac{dx}{y}=2i\int_{-1}^1\frac{dx}{\sqrt{1-x^2}},$$because each copy of the integral corresponds to one of two semicircular arcs.${}^\dagger$ This is$$2i[\arcsin1-\arcsin(-1)]=4i\arcsin1=2i\pi.$$
${}^\dagger$ The reason it's $2i$ rather than $-2i$ is because by definition, $\oint$ goes anticlockwise. Going from $\theta=0$ to $\theta=\pi$ runs from $x=1$ to $x=-1$ with $y=\sqrt{1-x^2}$; going from $\theta=\pi$ to $\theta=2\pi$ runs from $x=-1$ to $x=1$ with $y=-\sqrt{1-x^2}$. Therefore, both arcs also give a sign change.