$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\ =\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.
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 $$\displaystyle I = \frac{\pi}{8}\ln 2+\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \theta d\theta-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta = \frac{\pi}{8}\ln 2$$
Let $x+1=\tan y$
$$I=\int_0^{\infty}\dfrac{\ln x}{(x+1)^2+1}dx=\int_{\frac\pi4}^{\frac\pi2}\ln(\tan y-1)dy$$
Using $\displaystyle\int_a^bf(y)\ dy=\int_a^bf(a+b-y)\ dy,$
$$I=\ln2\int_{\frac\pi4}^{\frac\pi2}dy-I$$
Hope you can take it from here!
$\bf{Another\; Solution::}$ 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 $$\displaystyle I = \frac{\ln 2}{2} \times \left[\frac{\pi}{2}-\frac{\pi}{4}\right] = \frac{\pi}{8}\cdot \ln 2$$
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 $\epsilon\rightarrow 0.$
