Residue Theorem complex analysis

Question

$$ \int_{-\infty}^{+\infty} \frac{2i \sqrt[3]x + e^{2ix}}{x^2+4}$$ .

I want to calculate integral for this . I know I have to use residue theorem to get simplified version of this but How to do it?

Answer 1

Observe first that $$ \frac{\sqrt[3]{x}}{x^2+4} $$ is an odd function, and hence $$\int_{-\infty}^\infty\frac{\sqrt[3]{x}\,dx}{x^2+4}=0.$$

Also, $$\frac{\sin 2x}{x^2+4},$$ is an odd function. Hence, it remains to calculate $$ \int_{-\infty}^\infty \frac{\cos 2x\,dx}{x^2+4}. $$

See relative post and also.