If C is the arc of the circle $|z|=1$ that lies in the first quadrant, then $$\left|\int_C \operatorname{Log}(z)dz\right|\leq \frac{\pi^2}{4}.$$
The trick I think is to use the triangle inequality for $|\operatorname{Log}(z)|\leq |\operatorname{Log}|z||+|i\operatorname{Arg}(z)|$, but I'm stuck after that.
You're almost done. $\newcommand{\Log}{\operatorname{Log}}\newcommand{\Arg}{\operatorname{Arg}}$
\begin{align} \left| \int_C \operatorname{Log} z\,dz \right| &\le \max_{z\in C} \left| \Log z \right| \cdot \ell(C) \\ &= \max_{z\in C} \left| \ln|z| + i\Arg z\right| \cdot \ell(C) \\ &= \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4} \end{align}
(assuming you mean the principal branch of the complex logarithm). Note that $ln|z| = 0$ for $z \in C$ and that $0 \le \Arg z \le \frac{\pi}2$. Also $\ell(C)$ is the length of $C$.