Seeking methods to solve $\int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$ I recently asked for definite integrals that can be solved using the Feynman Trick. One of the responses is the following integral:
$$I = \int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$$
I employed Laplace Transforms with Feynman's Trick (as per a response below) to solve it, but am curious about other approaches that can be taken (in particular if using Feynman's Tricks). 

My approach:
Let 
$$I(t) = \int_{0}^{\infty} \frac{xt - \sin(xt)}{x^3\left(x^2 + 4\right)} \:dx$$
Where $I = I(1)$
Taking the first derivative:
$$ \frac{dI}{dt} =  \int_{0}^{\infty} \frac{x - x\cos(xt)}{x^3\left(x^2 + 4\right)} \:dx  =  \int_{0}^{\infty} \frac{1 - \cos(xt)}{x^2\left(x^2 + 4\right)} \:dx$$
Taking the second derivative:
$$ \frac{d^2I}{dt^2} =  \int_{0}^{\infty} \frac{x\sin(xt)}{x^2\left(x^2 + 4\right)} \:dx  =  \int_{0}^{\infty} \frac{\sin(xt)}{x\left(x^2 + 4\right)} \:dx$$
Now, take the Laplace Transform w.r.t $t$:
\begin{align}
 \mathscr{L}\left[ \frac{d^2I}{dt^2}\right]   &=  \int_{0}^{\infty} \frac{\mathscr{L}\left[\sin(xt)\right]}{x\left(x^2 + 4\right)} \:dx \\
&=  \int_{0}^{\infty} \frac{x}{\left(s^2 + x^2\right)x\left(x^2 + 4\right)}\:dx \\
&= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx
\end{align}
Applying the Partial Fraction Decomposition we may find the integral
\begin{align}
\mathscr{L}\left[ \frac{d^2I}{dt^2}\right] &= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx \\
&= \frac{1}{s^2 - 4} \int_{0}^{\infty} \left[\frac{1}{x^2 + 4} - \frac{1}{x^2 + s^2} \right]\:dx \\
&= \frac{1}{s^2 - 4} \left[\frac{1}{2}\arctan\left(\frac{x}{2}\right) - \frac{1}{s}\arctan\left(\frac{x}{s}\right)\right]_{0}^{\infty} \\
&=  \frac{1}{s^2 - 4} \left[\frac{1}{2}\frac{\pi}{2} - \frac{1}{s}\frac{\pi}{2}  \right] \\
&= \frac{\pi}{4s\left(s + 2\right)}
\end{align}
We now take the inverse Laplace Transform:
$$ \frac{d^2I}{dt^2} = \mathscr{L}^{-1}\left[\frac{\pi}{4s\left(s + 2\right)} \right] = \frac{\pi}{8}\left(1 - e^{-2t} \right) $$
We now integrate with respect to $t$:
$$ \frac{dI}{dt} =  \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) + C_1$$
Now 
$$ \frac{dI}{dt}(0) = \int_{0}^{\infty} \frac{1 - \cos(x\cdot 0)}{x^2\left(x^2 + 4\right)} \:dx  =  0 = \frac{\pi}{8}\left(\frac{1}{2} \right) + C_1 \rightarrow C_1 = -\frac{\pi}{16}$$
Thus,
$$ \frac{dI}{dt} =  \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16}$$
We now integrate again w.r.t $t$
$$ I(t) = \int \left[\frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16} \right] \:dt = \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + C_2 $$
Now 
$$I(0) = \int_{0}^{\infty} \frac{x\cdot0 - \sin(x\cdot0)}{x^3\left(x^2 + 4\right)} \:dx = 0 = \frac{\pi}{8}\left( -\frac{1}{4} \right)  + C_2 \rightarrow C_2 = \frac{\pi}{32}$$
And so we arrive at our expression for $I(t)$
$$I(t)= \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + \frac{\pi}{32}$$
Thus, 
$$I = I(1) = \frac{\pi}{8}\left(\frac{1}{2} - \frac{e^{-2}}{4} \right) - \frac{\pi}{16} + \frac{\pi}{32} = \frac{\pi}{32}\left(1 - e^{-2}\right)$$
