$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of 
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
 A: Let $I$ be defined by the integral 
$$I=\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx \tag1$$
and let $J$ be the contour integral 
$$J=\oint_{C}\frac{\log^2(z)}{z^2+2z+4}\,dz \tag2$$
where the contour $C$ is the classical "key-hole" contour for which the keyhole coincides with the branch cut along the positive real axis.  
Then, we can write $(2)$ as
$$\begin{align}
J&=\int_0^R \frac{\log^2(x)}{x^2+2x+4}\,dx-\int_0^R \frac{(\log(x)+i2\pi)^2}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\\\\
&=-i4\pi \int_0^R \frac{\log(x)}{x^2+2x+4}\,dx+4\pi^2\int_0^R\frac{1}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\tag 3
\end{align}$$
As $R\to \infty$ the integral over $C_R$ vanishes and we have from $(1)$
$$\lim_{R\to \infty}J=-i4\pi I +4\pi^2\int_0^\infty\frac{1}{x^2+2x+4}\,dx\tag 4$$

Next, we apply the residue theorem to evaluate the left-hand side of $(4)$.  Proceeding we find that 
$$\begin{align}
\lim_{R\to \infty}J&=2\pi i \text{Res}\left(\frac{\log^2(z)}{z^2+2z+4}\,dz, z=-1\pm i\sqrt{3}\right)\\\\
&=2\pi i \left(\frac{(\log(2)+i2\pi/3)^2}{i2\sqrt{3}}+\frac{(\log(2)+i4\pi/3)^2}{-i2\sqrt{3}}\right)\\\\
&=-\frac{i4\pi^2\log(2)}{3\sqrt{3}}+\frac{4\pi^3}{3\sqrt{3}}\tag 5
\end{align}$$
Equating real and imaginary parts of $(3)$ and $(5)$ reveals

$$\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx =\frac{\pi \log(2)}{3\sqrt{3}}$$

and as a bonus
$$\int_0^\infty\frac{1}{x^2+2x+4}\,dx=\frac{\pi}{3\sqrt{3}}$$
And we are done!
A: Let $$I = \int^{\infty}_{0}\frac{\ln x}{x^2+2x+4}dx = \int^{\infty}_{0}\frac{\ln x}{(x+1)^2+(\sqrt{3})^2}dx$$
put $\displaystyle x=\frac{1^2+(\sqrt{3})^2}{y} = \frac{4}{y}$ and $\displaystyle dx = -\frac{4}{y^2}dy$
$$I = \int^{0}_{\infty}\frac{\ln (4)-\ln (y)}{16+8y+4y^2}\cdot -\frac{4}{y^2}\cdot y^2 dy = \int^{\infty}_{0}\frac{\ln 4-\ln y}{y^2+2y+4}dy$$
$$I = \ln 4\int^{\infty}_{0}\frac{1}{(y+1)^2+(\sqrt{3})^2}dy-I$$
So $$I = \ln 2 \cdot \frac{1}{\sqrt{3}}\cdot \tan^{-1}\left(\frac{y+1}{\sqrt{3}}\right)\bigg|^{\infty}_{0} = \frac{\ln 2}{\sqrt{3}}\cdot \bigg(\frac{\pi}{2}-\frac{\pi}{6}\bigg) = \frac{\pi \cdot \ln 2}{3\sqrt{3}}$$ 
A: May be this will be a more elementary solution.
Let us start with a the substitution $x=2y$. Then
\begin{gather*}
I = \int_{0}^{\infty}\dfrac{\ln(x)}{x^2+2x+4}\, dx = \int_{0}^{\infty}2\dfrac{\ln(2y)}{4y^2+4y+4}\, dy =\\[2ex] \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln(2)}{y^2+y+1}\, dy + \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln y}{y^2+y+1}\, dy=
\dfrac{\ln(2)}{2}I_1+ \dfrac{1}{2}I_2\tag{1}
\end{gather*}
where
\begin{equation*}
I_1 = \int_{0}^{\infty}\dfrac{1}{y^2+y+1}\, dy =\dfrac{4}{3}\int_{0}^{\infty}\dfrac{1}{\left(\frac{2y+1}{\sqrt{3}}\right)^{2}+1}\, dy = \dfrac{2}{\sqrt{3}}\left[\arctan\left(\dfrac{2y+1}{\sqrt{3}}\right)\right]_{0}^{\infty} = \dfrac{2\pi}{3\sqrt{3}}\tag{2}
\end{equation*}
and
\begin{equation*}
I_2 = \int_{0}^{\infty}\dfrac{\ln(y)}{y^2+y+1}\, dy =\left[y=\dfrac{1}{z}\right] = \int_{0}^{\infty}\dfrac{-\ln(z)}{z^2+z+1}\, dz = -I_2. \tag{3}
\end{equation*}
But (3) implies that $I_2=0$. This and (2) substituted in (1) give us $I = \dfrac{\pi\ln(2)}{3\sqrt{3}}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\quad r \equiv -1 + \root{3}\ic\quad\mbox{and}\quad  R > 0}$,

Hereafter, $\ds{\ln\pars{z}}$ is the principal $\ds{\ln}$-branch cut.
\begin{align}
\int_{0}^{4R}{\ln\pars{x} \over x^{2} + 2x + 4} & =
\int_{0}^{4R}\ln\pars{x}\pars{{1 \over x - r} - {1 \over x - \bar{r}}}
{1 \over r - \bar{r}}\,\dd x
\\[5mm] & =
{\root{3} \over 3}\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x
\label{1}\tag{1}
\end{align}

Then,
\begin{align}
\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x & =
-\,\Im\int_{0}^{4R}{\ln\pars{r\bracks{x/r}} \over 1 - x/r}\,{\dd x \over r} =
-\,\Im\int_{0}^{4R/r}{\ln\pars{rx} \over 1 - x}\,\dd x =
-\,\Im\int_{0}^{R\bar{r}}{\ln\pars{rx} \over 1 - x}\,\dd x
 \\[5mm] & =
-\,\Im\bracks{-\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} +
\int_{0}^{R\bar{r}}{\ln\pars{1 - x} \over x}\,\dd x}
\\[5mm] & =
-\,\Im\bracks{-\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} -
\,\mrm{Li}_{2}\pars{R\,\bar{r}}}
\end{align}

With $\quad$ $\ds{\,\mrm{Li}_{2}}$-Inversion Formula$\quad$ and $\ds{\quad R\,\bar{r} \not\in \left[0,1\right)}$:
\begin{align}
\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x & =
\Im\braces{\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} +
\bracks{-\,\mrm{Li}_{2}\pars{1 \over R\,\bar{r}} - {\pi^{2} \over 6} -
{1 \over 2}\,\ln^{2}\pars{-R\bar{r}}}}
\\[5mm] & =
\Im\braces{\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} -
{1 \over 2}\,\ln^{2}\pars{-R\,\bar{r}}
- \,\mrm{Li}_{2}\pars{1 \over R\,\bar{r}}}
\end{align}

\begin{align}
&\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x
\\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\,&
\Im\braces{\ln\pars{1 + R + R\root{3}\ic}\ln\pars{4R} -
{1 \over 2}\,\ln^{2}\pars{R + R\root{3}\ic}}
\\[5mm] = &\
\arctan\pars{{R \over 1 + R}\root{3}}\ln\pars{4R} -
{1 \over 2}\,\Im\pars{\bracks{%
\ln\pars{2R} + \arctan\pars{\root{3}}\ic}^{2}}
\\[5mm] = &\
\arctan\pars{{R \over 1 + R}\root{3}}\ln\pars{4R} -
\ln\pars{2R}\,\arctan\pars{\root{3}}
\\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\to}\,\,\,&
\bbx{\ds{{1 \over 3}\,\pi\ln\pars{2}}}\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:
$$
\bbx{\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 2x + 4} =
{\root{3} \over 9}\,\pi\ln\pars{2}}} \approx 0.4191
$$
