$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.

Integrate  $$f(z)  =  \frac{\log^2  z}{(z+1)^2+1}$$  along  a  keyhole
contour with  the branch cut of the  logarithm on the positive  real axis and
its argument between $0$ and $2\pi.$
The poles are at $$\rho_{1,2} = -1 \pm i$$ with residues
$$\frac{\log^2 \rho_{1,2}}{2\rho_{1,2}+2}$$
and these sum to
$$\frac{(1/2 \log 2 + 3/4\pi i)^2}{2i}
- \frac{(1/2 \log 2 + 5/4\pi i)^2}{2i}
\\ = \frac{1}{2i} (-1/2 (\log 2) \pi i + \pi^2)
= - \frac{1}{4} (\log 2) \pi
- \frac{1}{2} \pi^2 i$$
for a contribution of
$$2\pi i \times
\left(- \frac{1}{4} (\log 2) \pi
- \frac{1}{2} \pi^2 i\right)
= -\frac{1}{2} (\log 2) \pi^2 i
+ \pi^3.$$
The contribution from the circular  arc vanishes and on the lower part
of the  key slot  we obtain  after cancellation of  the square  of the
logarithm 
$$-\int_0^\infty \frac{2\times 2\pi i \log x - 4\pi^2}{x^2+2x+2} dx.$$
Equating real and complex parts we obtain the two integrals
$$-4\pi \int_0^\infty \frac{\log x}{x^2+2x+2} dx
= -\frac{1}{2} (\log 2) \pi^2$$
or
$$\int_0^\infty \frac{\log x}{x^2+2x+2} dx 
= \frac{1}{8} (\log 2) \pi$$
and
$$4\pi^2 \int_0^\infty \frac{1}{x^2+2x+2} dx
= \pi^3$$
or
$$\int_0^\infty \frac{1}{x^2+2x+2} dx = \frac{1}{4} \pi.$$
We may verify by ML that on  the large circle say of radius $R$ we get
the estimate of the modulus of the integral
$$2\pi R \times \frac{\log^2 R}{R^2}
\rightarrow 0$$
as $R\rightarrow\infty.$
We get for the small circle of radius $\epsilon$ encircling the origin
$$2\pi \epsilon \times \frac{\log^2 \epsilon}{2}
\sim \pi \epsilon \times \log^2 \epsilon
\rightarrow 0$$
as $\epsilon\rightarrow 0.$
