Prove $\mid \int_\ell \frac{z^3}{z^2+1}dz \mid \leq \frac{9\pi}{8}$ where $\ell$ is $|z|=3,\Re(z)\geq 0$.
My attempt: $$\mid \int_\ell \frac{z^3}{z^2+1}dz \mid \leq \int_\ell \frac{\mid z\mid^3}{\mid z^2+1\mid} \ dz $$ and therefore, since $|z|=3\Rightarrow |z|^3=27, |z^2+1|\geq |z|^2-1=8$, we obtain $$ \mid \int_\ell \frac{z^3}{z^2+1}dz \mid \leq \frac{27}{8}\cdot 3\pi, $$ where $3\pi $ is the curve perimeter. The same minimal value for the denominator is obtained when taking $z=x+iy$ and examining the value of $|z^2+1|^2 = 4\cdot (25-y^2)$ (where $x=\sqrt{9-y^2}$).
Unfortunately, I cannot see how I can refine this argument to obtain the required bound. Will be happy to receive some help with this.
Thank you
The trick here is to reduce the power by polynomial division, so $\frac{z^3}{z^2+1}=z-\frac{z}{z^2+1}$
Then $$|\int_\ell \frac{z^3}{z^2+1}dz|= |-\int_\ell \frac{z}{z^2+1}dz +\int_\ell z dz| $$
By direct computation or Cauchy (moving to the vertical segment from $-3$ to $3$) we have $\int_\ell z dz=0$, so now your reasoning gives the required bound $9\pi/8$