<s>[INST] $\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.
 [/INST] Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \int_{0}^{\infty}\frac{\ln x}{(x+1)^2+1^2}dx$$
Put $(x+1) = \tan \theta \;,$ Then $dx = \sec^2 \theta d\theta$  and changing Limits
We get $$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\ln\left(\tan \theta - 1\right)}{1+\tan^2 \theta }\cdot \sec^2 \theta d\theta  = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \left(\sin \theta -\cos \theta \right)d\theta -\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta$$
So $$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left(\sqrt{2}\cdot \sin \left(\theta-\frac{\pi}{4}\right)\right)d\theta-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\cos \theta d\theta$$
$$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\sqrt{2}d\theta+\int_{0}^{\frac{\pi}{4}}\ln\sin \left(\theta\right)d\theta -\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta $$
Now In Second Integral $\displaystyle \theta = \left(\frac{\pi}{2}-\phi\right)\;,$ then $d\theta = -d\phi$ and Changing Limit
We Get $$\displaystyle I = \frac{\pi}{8}\ln 2-\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \phi d\phi-\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \theta d\theta$$
So $$\displaystyle I = \frac{\pi}{8}\ln 2+\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \theta d\theta-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta = \frac{\pi}{8}\ln 2$$
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