How to show that $\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?$ How to show that $(1)$ is 
$$\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?\tag1$$
$$x^2+3x+3=\left(x+{3\over 2}\right)^2+{3\over 4}$$
$$\int_{0}^{\infty}e^{-x}\cos(x){\mathrm dx\over (1+x)^3}+\int_{0}^{\infty}e^{-x}\left(x+{3\over 2}\right)^2\cos(x){\mathrm dx\over (1+x)^3}=I_1+I_2\tag2$$
$$(1+x)^{-3}=\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}x^n$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\cos(x)\mathrm dx=I_1\tag3$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\left(x+{3\over 2}\right)^2\cos(x)\mathrm dx=I_2\tag4$$

Start with partial fractions:
$$ \frac{x^2+3x+3}{(1+x)^3} = \frac{1}{1+x} + \frac{1}{(1+x)^2} + \frac{1}{(1+x)^3} $$
Now let
$$ J_k = \int_0^\infty e^{-(1+i)x}\frac{dx}{(1+x)^k} $$
Integrating by parts,
$$ J_k = \frac{1-i}{2} - \frac{((1-i)k}{2} J_{k+1} $$
Using this for $k=1$ and $k=2$,  $J_1 + J_2 + J_3$ simplifies to $1 - i/2$.
Your integral is the real part of this, namely $1$. 
