Use the principal branch of log z $\int_{-1}^{1} log z\, dz$  Use the principal branch of log z
$\int_{-1}^{1} log z\, dz$ 
My attempt was
we see how -1 to 1 became pi to 0? That should make the e^whatever terms go away. Then we can do it by parts.
$\int_{-1}^{1} log z\, dz$ 
=$\int_{\pi}^{0} log (e^{i\theta})ie^{i\theta}\, dz=-2+i\pi$
Is this correct result?
Could please show me another method of resolution?
Could someone help me through this problem?

I would say
$$
\int_{-1}^1 \log(z)\,dz = \int_0^1\log(x)\,dx+\int_{-1}^0[\log(-x)+i\pi]\,dx
$$
and then do some real integrals.
