Show that
$$\int_C \frac{2 z^2-5}{(z^2+1)(z^2+4)} dz \le \frac{\pi R (2 R^2+5)}{(R^2-1)(R^2-4)} $$
Let $C$ be the upper half of the circle $z=R$ for any $R>2$.
Do I need to actually find the integral here? Can someone help me start this question please?
Actually, the question should read
$$\left |\int_C \frac{2 z^2-5}{(z^2+1)(z^2+4)} dz\right | \le \frac{\pi R (2 R^2+5)}{(R^2-1)(R^2-4)} $$
Here, you may use the ML-lemma/theorem, which states that the magnitude of an integral over a contour is less than or equal to the maximum value of the modulus of the integrand times the length of the contour. The length in this case is $\pi R$. The max modulus clearly occurs for $z=i R$. The RHS of the inequality follows.