Cauchy Principal Value Integral calculation

Question

How can i resolve this integral in Cauchy principal value? $$\int_{-\infty}^ \infty \! \frac{x+\sin x}{x(x^2+4j-4)^2} \ \mathrm{d}x $$ Then $$\int_{-\infty}^ \infty \! \frac{1}{(x^2+4j-4)^2} \ \mathrm{d}x + \frac {1}{2j} \int_{-\infty}^ \infty \! \frac{e^{jz}-e^{-jz}}{x(x^2+4j-4)^2} \ \mathrm{d}x $$ I usually resolve them with Jordan's lemmas, easily applicable to the second one, but for the first one?

Answer 1

$$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\ub}{\underbrace} \newcommand{\t}{\text} \newcommand{\s}{\sqrt} I=\int_{-\infty}^{\infty}\frac{x+\sin x}{x(x^2+4i-4)^2}{\rm d}x=\ub{\int_{-\infty}^{\infty}\frac{1}{(x^2+4i-4)^2}{\rm d}x}_{I_1}+\ub{\int_{-\infty}^{\infty}\frac{\sin x}{x(x^2+4i-4)^2}{\rm d}x}_{I_2}\\ \t{Now:} \\I_1=\int_{-\infty}^{\infty}\frac{1}{({x-2\s{1-i}})^2(x+2\sqrt{1-i})^2}{\rm d}x\\ \t{Now taking a disc in positive imaginary argand plane and integrating$\\$ around the contour along with taking residue at the root in the upper plane:}\\ \int_{-\infty}^{\infty}\frac{1}{({x-2\s{1-i}})^2(x+2\sqrt{1-i})^2}{\rm d}x=2\pi i\b{\frac1{64}i\s{1-i}}=-\frac{\pi}{32}\s{1-i}$$