What is the correct way to do $\int_{-1}^1 \frac {dx}{x+1+2i} $?

Question

The question is to evaluate

$ \int_{-1}^{1} \frac {dx}{x+1+2i} $

My friend suggested

$ \int_{-1}^{1} \frac {dx}{x+1+2i} $

$= \int_{-1}^{1} \frac {d(1+x+2i)}{x+1+2i} $

$ =\big[\ln (1+x+2i)\big]_{-1}^1 \quad (*)$

$=\ln \big( \frac{2+2i}{2i} \big)$

$=\ln \big(\frac{1+i}i\big)$

$=\ln (1-i)$

$=\frac 12 \ln2-i\frac{\pi}4$

which is actually the correct answer !

I raised objection in the step $(*)$ saying

" The anti-derivative of $\frac 1z$ is not $\ln (z)$ since if this was the case the integral of $\frac 1z$ over any closed curve containing zero would have been zero which is contrary ."

Instead I suggested to multiply the numerator and denominator by the conjugate of the denominator and separately integrate the real and imaginary parts ( which gives same answer.)

I don't think I had convinced him (He is from Applied Maths background) , but I want to know if I am correct on my side. If not, then what is the correct explanation.

Thanks for the time !

Answer 1

This integration is still one dimensional where constants are non-real. So the ideas of contour and integration over $z(x+iy)$ is not meant here. So, both the approach and the answer are right.