Question: 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 .

Answer: 
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 $I = \dfrac{\pi\ln(2)}{3\sqrt{3}}$.
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 $$I = \ln 2 \cdot \frac{1}{\sqrt{3}}\cdot \tan^{-1}\left(\frac{y+1}{\sqrt{3}}\right)\bigg|^{\infty}\_{0} = \frac{\ln 2}{\sqrt{3}}\cdot \bigg(\frac{\pi}{2}-\frac{\pi}{6}\bigg) = \frac{\pi \cdot \ln 2}{3\sqrt{3}}$$