Seeking methods to solve $ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$ I am currently working on an definite integral that requires the following definite integral to be evaluated. 
$$ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$$
Where $k \in \mathbb{R}^{+}$
I was wondering what methods can be employed to solve this integral?

Let's take the advice from the comment section and evaluate the integral using contour integration! Consider the function$$f(z)=\frac {e^{iaz}}{z(1+z^2)}$$And integrate it about the semi-circular contour situated in the upper-half of the complex plane with a small semi circular detour around the origin. Let the larger circle have a radius of $R$ while the smaller detour $\epsilon$ (Picture coming soon). Therefore, calling the contour $\mathrm C$, we have that$$\oint\limits_{\mathrm C}\mathrm dz\, f(z)=\int\limits_{-R}^{-\epsilon}\mathrm dx\, f(x)+\int\limits_{\gamma_{\epsilon}}\mathrm dz\, f(z)+\int\limits_{\epsilon}^{R}\mathrm dx\, f(x)+\int\limits_{\Gamma_{R}}\mathrm dz\, f(z)$$The arc integrals obtain different values depending on the radius. The outer arc integral vanishes per the Estimation Lemma as$$\left|\,\int\limits_{\Gamma_{R}}\mathrm dz\, f(z)\,\right|\leq\frac {\pi R}{R(R^2-1)}\xrightarrow{R\,\to\,\infty}0$$Meanwhile$$\left|\,\int\limits_{\epsilon_{\epsilon}}\mathrm dz\, f(z)\,\right|=-i\int\limits_0^{\pi}\mathrm d\varphi\,\frac {e^{ia\epsilon e^{i\varphi}}}{1+\epsilon^2e^{2i\varphi}}\xrightarrow{\epsilon\,\to\,0}-\pi i$$The contour integral is equal to $2\pi i$ times the sum of the residues. There is only one pole inside our contour at $z=i$. The corresponding residue is calculated as$$\operatorname*{Res}_{z\, =\, i}\frac {e^{iaz}}{z(1+z^2)}=\lim\limits_{z\,\to\, i}\frac {(z-i)e^{iaz}}{z(1+z^2)}=-\frac 1{2e^a}$$Hence$$\oint\limits_{\mathrm C}\mathrm dz\, f(z)\color{red}{=-\pi i e^{-a}}$$Substituting all the values back into the original equation gives$$\int\limits_{-\infty}^{\infty}\mathrm dx\, f(x)=\pi i(1-e^{-a})$$Take the imaginary part of the integral to get$$\int\limits_0^{\infty}\mathrm dx\,\frac {\sin ax}{x(1+x^2)}\color{blue}{=\frac \pi2\left(1-\frac 1{e^a}\right)}$$
