I am trying to find the value of:
$$\int_\Gamma \frac{1}{1+z^2} dz.$$
where $\Gamma$ is the straight line, going from $1$ to $1+i$ my plan was to use the parameterization $\gamma(t)=1+it$. So $\gamma(t)'=i$. Now using the formula:
$$\int_\Gamma f(z) \, dz= \int_0^1 f(\gamma(t))\gamma(t)' \, dt$$
$$\int_\Gamma \frac{1}{1+z^2} dz= \int_0^1 \frac{i}{1+(1+it)^2} \, dt$$ $$\int_\Gamma \frac{1}{1+z^2} dz= \int_0^1 \frac{i}{2+2it-t^2} \, dt$$
However I can't seem to get much further, I know you could use partial fractions and then use logs but I was trying to do it a different method!
You have\begin{align}\int_0^1\frac i{2+2it-t^2}\,\mathrm dt&=\int_0^1\frac{i(2-2it-t^2)}{(2+2it-t^2)(2-2it-t^2)}\,\mathrm dt\\&=\int_0^1\frac{2t}{4+t^4}\,\mathrm dt+\left(\int_0^1\frac{2-t^2}{4+t^4}\,\mathrm dt\right)i\\&=\frac\pi4-\frac{\arctan2}2+\frac{\log5}4i.\end{align}