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

$\bf{Another\; Solution::}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx.$$
Put $\displaystyle x = \frac{2}{t}\;,$ Then $\displaystyle dx = -\frac{2}{t^2}dt$ and Changing Limits, We get
$$\displaystyle I = \int_{\infty}^{0}\frac{\ln\left(\frac{2}{t}\right)}{\frac{4}{t^2}+\frac{4}{t}+2}\cdot -\frac{2}{t^2}dt = \int_{0}^{\infty}\frac{\left(\ln 2-\ln t\right)}{t^2+2t+2}dt$$
Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$ and $$\displaystyle \bullet \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$
So $$\displaystyle I = \ln 2\int_{0}^{\infty}\frac{1}{x^2+2x+2}dx - \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \ln 2\int_{0}^{\infty}\frac{1}{(x+1)^2+1^2}dx-I$$
So we get $$\displaystyle I = \ln 2\times \left[\tan^{-1}\left(x+1\right)\right]_{0}^{\infty}-I$$
So we get $$\displaystyle I = \frac{\ln 2}{2} \times \left[\frac{\pi}{2}-\frac{\pi}{4}\right] = \frac{\pi}{8}\cdot \ln 2$$
