= Solution
The contour has <winding number> one around the single pole at zero. By the <Cauchy integral formula>, or the <residue> of $1/z$,
$$
\boxed{\oint_C\frac{dz}z=2\pi i.}
$$
Rotation by $z\mapsto iz$ carries each oriented edge to the next oriented edge and leaves $dz/z$ unchanged. The four edge integrals are therefore equal, each $\pi i/2$.
On the right edge use $z=1+it$, $-1\le t\le1$, so
$$
\frac{\pi i}{2}=\int_{-1}^1\frac{i\,dt}{1+it}
=\int_{-1}^1\frac{t+i}{1+t^2}\,dt
=i\int_{-1}^1\frac{dt}{1+t^2}.
$$
The real integrand $t/(1+t^2)$ is odd and integrates to zero. Dividing by $i$ gives \b[$\int_{-1}^1(1+t^2)^{-1}dt=\pi/2$] without needing its real antiderivative.
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