How to prove $\int_0^{\infty}\frac{x^2+3x+3}{(x+1)^3} e^{-x}\sin x\, dx = \frac{1}{2}.$ I've just seen this integral crop up on another site and I can't see an obvious way to prove it. 
Any suggestions?

$$\int_0^{\infty}\frac{x^2+3x+3}{(x+1)^3} e^{-x}\sin x\, dx = \frac{1}{2}.$$

 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}{x^{2} + 3x + 3 \over \pars{x + 1}^{3}}\,
     \expo{-x}\sin\pars{x}\,\dd x = \half:\ {\large ?}}$

With $\ds{a \equiv 1 - \ic}$:
\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}{x^{2} + 3x + 3 \over
\pars{x + 1}^{3}}\,\expo{-x}\sin\pars{x}
\,\dd x}
\\[5mm] = &\
\Im\int_{0}^{\infty}{1 + \pars{x + 1} + \pars{x + 1}^{2} \over \pars{x + 1}^{3}}\,
\expo{-ax}\,\dd x
\\[5mm] = &\
\sum_{n\ =\ 0}^{2}\Im\int_{0}^{\infty}
{\expo{-ax} \over \pars{x + 1}^{n + 1}}\,\dd x
\\ = &\
\sum_{n\ =\ 0}^{2}\Im\int_{0}^{\infty}
\expo{-ax}\,\ \overbrace{%
{1 \over n!}\int_{0}^{\infty}t^{n}\expo{-\pars{x + 1}t}\,\dd t}
^{\dsc{1 \over \pars{x +1}^{n + 1}}}\ \,\dd x
\\[5mm] = &\
\sum_{n\ =\ 0}^{2}{1 \over n!}\,\Im\int_{0}^{\infty}t^{n}\expo{-t}
\int_{0}^{\infty}\expo{-\pars{a + t}x}\,\dd x\,\dd t
\\[5mm] = &\
\sum_{n\ =\ 0}^{2}{1 \over n!}\,
\Im\int_{0}^{\infty}{t^{n}\expo{-t} \over a + t}\,\dd t
\\[5mm] = &\
\sum_{n\ =\ 0}^{2}{1 \over n!}\,
\int_{0}^{\infty}{t^{n}\expo{-t} \over \pars{t + 1}^{2} + 1}\,\dd t
\\[5mm] = &\
\half\int_{0}^{\infty}{%
2\sum_{n\ =\ 0}^{2}\,\,t^{n}/n! \over
\pars{t + 1}^{2} + 1}\,\expo{-t}\,\dd t
\\[5mm] = &\
\half\int_{0}^{\infty}
{2 + 2t+ t^{2}\over t^{2} + 2t + 2}\,
\expo{-t}\,\dd t =
\half\int_{0}^{\infty}\expo{-t}\,\dd t=
\color{#66f}{\Large\half}
\end{align}
A: Hint: Let $~I(a)=\displaystyle\int_0^\infty\frac{P(x)}{x+a}~e^{kx}~dx.~$ After applying polynomial long division to $~\dfrac{P(x)}{x+a}$ , 
evaluate $I''(1)$. Euler's formula will also come into play. P.S.: Do not be afraid of the exponential integrals and/or incomplete $\Gamma$ functions which will inevitably appear in the expression of $I(a)$: they will just as easily disappear when differentiating.
A: We have:
$$ I = \int_{0}^{+\infty}\frac{\sin x}{x}e^{-x}\,dx -\int_{0}^{+\infty}\frac{\sin x}{x(x+1)^3}e^{-x}\,dx=\color{blue}{I_1}-\color{red}{I_2}.$$
Since
$$\frac{\sin x}{x}=\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k}}{(2k+1)!}$$
and 
$$\int_{0}^{+\infty}x^{2k}e^{-x}\,dx = (2k)! $$
we have:
$$\color{blue}{I_1}=\int_{0}^{+\infty}\frac{\sin x}{x}e^{-x}\,dx = \sum_{k=0}^{+\infty}\frac{(-1)^k}{2k+1}=\arctan(1)=\color{blue}{\frac{\pi}{4}}.$$
Using a standard trick:
$$\int_{0}^{+\infty}\frac{\sin x}{x(x+1)^3}e^{-x}\,dx=\frac{1}{2}\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{\sin x}{x}e^{-x}t^2 e^{-t(x+1)}\,dt\,dx$$
and by the previous lemma we get:
$$\color{red}{I_2}=\int_{0}^{+\infty}\frac{\sin x}{x(x+1)^3}e^{-x}\,dx=\frac{1}{2}\int_{0}^{+\infty}t^2 e^{-t}\arctan\frac{1}{1+t}\,dt.$$
Now we use integration by parts. We have:
$$\color{red}{I_2} = \left.\frac{1}{2}e^{-t}(-t^2-2t-2)\arctan\frac{1}{t+1}\right|_{0}^{+\infty}-\frac{1}{2}\int_{0}^{+\infty}e^{-t}\,dt = \color{red}{\frac{\pi}{4}-\frac{1}{2}}$$
and we are done.
A: Somewhat more generally, let 
$$ F(s) = \int_0^\infty \dfrac{x^2+a x+ b}{(x+1)^3} e^{-sx}\; dx$$
which converges for $\text{Re}(s) > 0$.  It is the Laplace transform of
$$ f(x) = \dfrac{x^2 + a x + b}{(x+1)^3}$$
The general solution is
$$ F(s) = \dfrac{e^s}{2} \left((b-a+1) s^2 + (4-2a) s +2\right) Ei(1,s) + 
\left(a-1-b\right) \dfrac{s}{2}+\dfrac{a+b-3}{2} $$
where $$Ei(1,s) = \int_1^\infty \dfrac{e^{-st}}{t}\; dt$$
is the exponential integral function.
For an elementary solution, what you need is $(b-a+1) s^2 + (4 - 2 a) s + 2 = 0$
which is what you have in the case at hand with $a=b=3$, $s = 1 \pm i$.
EDIT: 
That "general solution" comes about this way.  First make the change of variables $x+1=t$, and expand $f(x)$ in powers of $t$.  Now as I mentioned
$$ \int_1^\infty \dfrac{e^{-st}}{t} \; dt = Ei(1,s)$$
Use integration by parts to get
$$ \eqalign{ \int_1^\infty \dfrac{e^{-st}}{t^3}\; dt &= \left. -{\frac {{{ e}^{-st}}}{2{t}^{2}}}\right|_{t=1}^\infty - \dfrac{s}2 \int_1^\infty {\frac {{{ e}
^{-st}}}{{t}^{2}}}\,{ d}t\cr
\int_1^\infty \dfrac{e^{-st}}{t^2}\; dt  &= \left. -{\frac {{{ e}^{-st}}}{{t}}}\right|_{t=1}^\infty - s \int_1^\infty {\frac {{{ e}
^{-st}}}{{t}}}\,{ d}t\cr}$$
