Prove that $\int_0^\infty\,\frac{\sin(kx)}{x(x^2+1)}\,\text{d}x=\frac{\pi}{2}\,\left(1-\exp(-k)\right)$ for all $k\in\mathbb{R}_{\ge0}$.

HINT:

Let $f(k)$ be given by the integral

$$f(k) =\int_0^\infty \frac{\sin(kx)}{x(x^2+1)}\,dx \tag1$$

Inasmuch as the improper integral $\int_0^\infty \frac{x\sin(kx)}{x^2+1}\,dx$ converges uniformly for $|k|\ge \delta>0$, we can differentiate twice under the integral in $(1)$ to reveal

$$f''(k)-f(k)=-\frac\pi2 \text{sgn}(k)\tag2$$

Solve $(2)$ subject to the initial conditions $f(0)=0$ and $f'(0)=\frac\pi2$.

NOTE:

$$\int_0^\infty \frac{x\sin(kx)}{x^2+1}\,dx=\int_0^\infty \frac{(x^2+1-1)\sin(kx)}{x(x^2+1)}\,dx=\int_0^\infty \frac{\sin(kx)}{x}\,dx-\int_0^\infty \frac{\sin(kx)}{x(x^2+1)}\,dx$$

You can use a nice property of the Laplace transform to computer your integral: $$\int_0^{\infty} f(x) g(x)\,dx=\int_0^{\infty} \mathcal{L}\{f(x)\}(s)\,\mathcal{L}^{-1}\{g(x)\}(s) \,ds$$ In your case, letting $f(x)=\sin (kx)$ and $g(x)=\frac{1}{x(x^2+1)}$, we can show that \begin{align} \int_0^{\infty} \frac{\sin(kx)}{x(x^2+1)}\,dx &= k\int_0^{\infty} \frac{1-\cos(s)}{k^2+s^2}\,ds \\ &=\frac{\pi}{2}-k\int_0^{\infty} \frac{\cos(s)}{k^2+s^2}\,ds \\ &=\frac{\pi}{2}-\frac{k}{2}\int_{-\infty}^{\infty} \frac{e^{is}}{k^2+s^2}\,ds \\ &=\frac{\pi}{2} - \frac{k}{2} \sqrt{2\pi} \,\mathcal{F}\left\{\frac{1}{k^2+s^2}\right\}(\omega)\Biggr|_{\omega=1} \\ &= \frac{\pi}{2} - \frac{k}{2}\sqrt{2\pi} \sqrt{\frac{\pi}{2}} \frac{e^{-k}}{k} \\ &=\frac{\pi}{2}-\frac{\pi}{2}e^{-k} \\ &= \frac{\pi}{2}\left(1-e^{-k}\right) \end{align} Where we used the Fourier Transform. Thus, $$\int_0^{\infty} \frac{\sin(kx)}{x(x^2+1)}\,dx=\frac{\pi}{2}\left(1-e^{-k}\right) \quad \text{for } k\in\mathbb{R}^+$$