Using the estimation lemma show that $$\left|\int\limits_C \frac{z^3}{z^2-1}\text{d}z\right|\le \frac{9}{8}\pi$$ where $C:\{z:|z|=3,\Re(z)\ge 0\}$.
The length of $C$ is $\pi$ and $\displaystyle \max_{z\in C}\left|\frac{z^3}{z^2-1}\right|\le\frac{3^3}{3^2-1}=\frac{27}{8}$, hence one get by estimation lemma that $$\left|\int\limits_{C}\frac{z^3}{z^2-1}\text{d}z\right|\le\frac{27}{8}\pi$$How can we show the required bound?
Since $$ \int_C z dz=\left[\frac{z^2}{2}\right]_{-3i}^{3i}=0,$$ we have \begin{align} \int_C \frac{z^3}{z^2-1}dz&=\int_C \left(z+\frac{z}{z^2-1}\right)dz\\ &=\int_C \frac{z}{z^2-1}dz.\end{align} Using the estimation lemma we have$$ \left| \int_C \frac{z}{z^2-1}dz\right|\le \int_C \frac{\,3\,}{8}|dz|=\frac{\,9\,}{8}\pi,$$ which implies $$ \left|\int_C \frac{z^3}{z^2-1}dz\right|\le \frac{\,9\,}{8}\pi.$$