$\displaystyle \int\limits_{1+i}^{2+i}s\, dx$
Where s is the complex conjugate of x
I have the answer but I don't have the foggiest idea how it is obtained.
answer: [(2^2 -1 )/2] -i
Define $\displaystyle x = e^{i\theta}$.
Then $\displaystyle \overline{x} = e^{-i\theta}$
and $\displaystyle dx = i\,e^{i\theta}$.
Also note that $\displaystyle \theta = \frac{\log{x}}{i}$.
So the integral becomes
$\displaystyle \int{\overline{x}\,dx} = \int{e^{-i\theta}\,i\,e^{i\theta}\,d\theta}$
$\displaystyle = \int{i\,d\theta}$
$\displaystyle = i\theta + C$
$\displaystyle = i\left(\frac{\log{x}}{i}\right) + C$
$\displaystyle = \log{x} + C$.
Now you should be able to substitute the limits of integration.
Remember that $\displaystyle \log{x} = \ln{|x|} + i\,\textrm{arg}\,{x}$.