Show that
$$ \left|\int_c \frac{e^{2z}}{1+e^z}\,dz \right|\leq \frac{2\pi e^{2R}}{e^{R}-1} $$
where $c$ is the vertical line segment from $z=R, (R>0)$ to $z=R+2\pi i $.
The answer tells me that the integral is less than $$\int_c \frac{e^{2x}}{e^x-1}\,dx$$ and then just evaluate it but I don't understand how they get there.
Hint: $|e^z|=e^{\text{Re}\,z}$ and $|e^z+1|\geq|e^z|-1=e^{\text{Re}\,z}-1$. Now apply some basic inequalities on the integrand using these estimates.