Evaluating $\int_0^\infty \frac{\ln x}{x^2+2x+2}dx$ $$\int_0^\infty \frac{\ln x}{x^2+2x+2}dx$$
A question similar to this one has been asked before here, but there the quadratic doesn't have a constant term.
I tried to use completing the square and use the answers in the linked question but since the log term in the numerator has $x$ not $x+1 $(after completing the square the quadratic is $(x+1)^2 + 1$) so that doesn't seem to work.
 A: Let's look at the complex valued function ($a<1$, so the function goes to zero as $|z|\rightarrow \infty$)
$$
f(z,a)=\frac{z^a}{z^2+2z+2} \quad (1)
$$
Furthermore  note that 
$$\partial_af(z,a)|_{a=0}= \frac{\log(z)}{z^2+2z+2} \quad (2)$$
Equating this expression at real positiv $x$ will give us the original integrand.
Now, let's integrate (1) around a contour in the complex plane, which looks like a keyhole with slit on the positive real axis. We get, applying the residue theorem
$$
\oint f(z,a) dz = \lim_{R \rightarrow \infty}\int_{C_R}f(z,a)dz+\lim_{\epsilon\rightarrow 0}\int_{C_\epsilon}f(z,a)dz+\int_0^{\infty}f(x+i\delta,a)dx+\int_{\infty}^0f(x-i\delta,a)dx=2\pi i(\text{res}(z=z_+)+\text{res}(z=z_-))
$$
where $z_{\pm}=-1\pm i$ denotes the zeros of the denominator of $f(z,a)$ ,$C_R$ is a big circle in the complex plane terminating at the branch cut and $C_{\epsilon}$ is a small semicircle around the origin (branch point).
Note that the first two integrals   will vanish after taking the limits (left as excersise). 
Our choice of contour implies the following definition of the complex logarithm ($x>0$):
$$
\lim_{\delta\rightarrow 0_+}\log(x+i\delta)=\log(x), \quad
\lim_{\delta\rightarrow 0_+}\log(x-i\delta)=\log(x)+2 \pi i
$$ 
implying 
$$
\oint f(z,a) dz =\\(1-e^{ 2\pi i a})\int_{0}^{\infty}f(x,a)dx=2\pi i(\text{res}(z=z_+)+\text{res}(z=z_-)) \quad (3)
$$
Calculating the residues is quiet easy if one is careful with the argument of $z^a$
$$
\text{res}(z=z_+)=-\frac{(\sqrt{2})^{a}e^{3\pi i a /4}}{2 i}, \quad
\text{res}(z=z_-)=\frac{(\sqrt{2})^{a}e^{5\pi i a/4}}{2 i}
$$
and (3) becomes
$$
\int_{0}^{\infty}f(x,a)dx= \pi (\sqrt{2})^a \frac{\sin(\pi a /4)}{\sin(a \pi)}
$$
Furthermore (using  (2))

$$I= \partial_a \int_{0}^{\infty}f(x,a)dx\big|_{a=0}=\frac{\pi}{8}\log(2)$$ 

is the integral inquestion.
Appendix
As a bonus we can obtain generic integrals of the type
$$
I(m,a)=\int_0^{\infty}dx\frac{x^a \log^m(x)}{x^2+2x+2}
$$
by just observing
$$
I(m,a)=\partial_m\int_0^{\infty}dx\frac{ x^a }{x^2+2x+2}
$$
and use the results above
A: I tried to use $x+1 = \tan t$ then we have
$$
I = \int^{\pi/2}_{\pi/4} \ln (\tan t-1) dt 
$$
We can use
$$
\int_a^b f(x)dx = \int_a^b f(a+b-x) dx
$$
so we have
$$
I = \int^{\pi/2}_{\pi/4} \ln (\tan t-1) dt = \int^{\pi/2}_{\pi/4} \ln (\tan \left(\frac{\pi}{4}+\frac{\pi}{2} - t\right)-1) dt 
$$
but we know
$$
\tan \left(\frac{\pi}{4}+\frac{\pi}{2} - t\right) = \frac{\tan\left(\frac{3\pi}{4}\right)-\tan t}{1+\tan\left(\frac{3\pi}{4}\right)\tan t}=\frac{-1-\tan t}{1-\tan t}
$$
so the argument for the last integral is
$$
\int^{\pi/2}_{\pi/4} \ln (\tan \left(\frac{\pi}{4}+\frac{\pi}{2} - t\right)-1) dt =\int^{\pi/2}_{\pi/4} \ln\left(\frac{-1-\tan t}{1-\tan t}-1\right)dt = \int^{\pi/2}_{\pi/4} \ln\left(\frac{1+\tan t}{\tan t-1}-\frac{\tan t-1}{\tan t-1}\right)dt
$$
or finally
$$
I = \int^{\pi/2}_{\pi/4} \ln\left(\frac{2}{\tan t-1}\right)dt = \int^{\pi/2}_{\pi/4}  \ln 2 dt - \int^{\pi/2}_{\pi/4} \ln (\tan t-1) dt  = \ln 2 \int^{\pi/2}_{\pi/4} dt - I
$$
so we get
$$
2I = \ln 2\left(\frac{\pi}{2}-\frac{\pi}{4}\right) = \frac{\pi \ln 2}{4}
$$
or finally
$$
I = \frac{\pi \ln 2}{8}
$$
A: Let $I$ be our integral. First, change of variable $x=2t$ gives:
$$I=\int_0^{+\infty}\frac{\log t}{2t^2+2t+1}+\log 2\int_0^{+\infty}\frac{dt}{2t^2+2t+1}$$
In the first integral, change of variable $t=1/u$:
$$\int_0^{+\infty}\frac{\log t}{2t^2+2t+1}=-\int_0^{+\infty}\frac{\log u}{u^2+2u+2}du=-I$$
Hence $$2I=\log 2\int_0^{+\infty}\frac{dt}{2t^2+2t+1}=\log(2)[{\rm Arctan}(2t+1)]_0^{+\infty}=\frac{\pi}{4}\log2$$
A: By completing the square the integral becomes : $$\int_{0}^{\infty}\frac{\log(x)}{(x+1)^2+1}$$Now do the change of variables $x+1=y$. The integral now becomes $$\int_1^{\infty } \frac{\log (y-1)}{y^2+1} \, dy$$ Now write $y=\frac{1}{z}$. We get $$\int_0^1 \frac{\log(1-z)-\log(z)}{z^2+1} \ dz$$ This integral can be written as $$\int_0^1 \frac{\log(1-z)-\log(1+z)+\log(1+z)-\log(z)}{z^2+1} \ dz$$ Further, $$\int_0^1 \frac{\log((1-z)/(1+z))}{z^2+1} \ dz -\int_0^1 \frac{\log(z)}{z^2+1}\ dz+\int_0^1 \frac{\log(1+z)}{z^2+1} \ dz$$ Now using the variable change $(1-z)/(1+z)=w$ for only the first integral . After this the first and the second integrals cancel out and we are left with the integral$$\int_0^1 \frac{\log(1+z)}{z^2+1} \ dz$$ This integral has been solved on MSE: Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$. Hence you get the integral. 
Note: A lot of this answer is owed to Chris's sis
