$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ 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 .

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!
