I'm doing the classical example of treating the integral $\int_{0}^{\infty }\frac{\log(x)}{\left (1+x^2 \right )^2}dx$ using residue theorem on the function $$f(z)=\frac{\log(z)}{\left (1+z^2 \right )^2}$$
I consider the contour $\gamma$ comprising two semi-circular arcs with radii $R$ and $\epsilon$ (having redefined the logarithm by deleting the negative imaginary axis).
I'm having trouble with estimating the part of $\gamma$ on the semi-circle $\left | z \right |=R$. I have:
$$\left |\int_{z\in \gamma ,\left | z \right |=R}f(z) \right |\leq \int_{0}^{\pi}\left | f(Re^{i\theta })iRe^{i\theta } \right |d\theta\leq \int_{0}^{\pi}R \frac{\sqrt{\log^{2}(R)+\theta ^{2}}}{(R^{2}-1)^{2}}d\theta $$
My book does the estimation:
$$\left | f(z) \right |\leq \frac{2\log(R)}{(R^{2}-1)^{2}}$$
How did they get the $2\log(R)$ estimation?
Thank's in advance.