Integrate $\int_0^\infty \frac{\log x}{x^2+2x+4}\ dx$ What is $$\int_0^\infty  \frac{\log x}{x^2+2x+4}\ dx$$ 
Here's a hint that came along with the problem,

Substitute $x$ as $2t$ and write it as a sum of two integrals. Then try to simplify.

 A: I will give it a try here. I also think this problem can be done without integration by parts, but the hint from the OP is very useful.
Using $x=2t$ we get
$\int_0^\infty \frac{\log x}{x^2+2x+4}\ dx$ = $2\int_0^\infty \frac{\log 2} {4t^2+4t+4}\ dt$ + $2\int_0^\infty \frac{\log t}{4t^2+4t+4}\ dt.$
The first integral:
Take the log term upfront and factor a $4$ from the denom. Complete the square on the denom and integrate accordingly to arrive at an arctan.
When you plug in values, you get $\frac{{\pi}log2}{\sqrt{27}}$ which is approx $0.18200$
It turns out that this is the answer to the integral. (With TI integrating from $0$ to $1000$ I came to $0.1787$)
Done? Of course not.
What about that second integral?
Let s call that integral $I$. Here is the trick: Perform a u-sub $t=\frac{1}{v}$
You can verify that you get essentially the same integral back with a negative coefficient upfront; that is $I=-cI$ from which it follows that $I=0.$
So I suspect this is the reason why you were given that hint, it is a very good one!
A: Another Approach
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^2+2x+4}\,\mathrm{d}x
&=\int_0^\infty\frac{\log(4)-\log(x)}{x^2+2x+4}\,\mathrm{d}x\tag{1}\\
&=\log(2)\int_0^\infty\frac{\mathrm{d}x}{x^2+2x+4}\tag{2}\\
&=\frac{\log(2)}{\sqrt3}\int_{1/\sqrt3}^\infty\frac{\mathrm{d}x}{x^2+1}\tag{3}\\
&=\frac{\pi\log(2)}{3\sqrt3}\tag{4}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto\frac4x$
$(2)$: average left and right sides of $(1)$
$(3)$: substitute $x\mapsto\sqrt3\,x-1$
$(4)$: arctan integral
A: To precise Imranfat answer,
$\displaystyle J=\int_0^\infty  \dfrac{\log x}{x^2+2x+4}\ dx$
Apply the change of variable $x=2t$,
$\displaystyle J=\int_0^\infty  \dfrac{\log (2t)}{2(t^2+t+1)}\ dt=\int_0^\infty  \dfrac{\log 2}{2(t^2+t+1)}dt+\int_0^\infty  \dfrac{\log t}{2(t^2+t+1)}dt$
In the right member the second integral is equal to zero,
$\displaystyle \int_0^\infty  \dfrac{\log (t)}{2(t^2+t+1)}\ dt=\int_0^1  \dfrac{\log (t)}{2(t^2+t+1)}\ dt+\int_1^\infty  \dfrac{\log (t)}{2(t^2+t+1)}\ dt$
Perform in the second integral the change of variable $y=\dfrac{1}{x}$, thus,
$\displaystyle \int_0^\infty  \dfrac{\log (t)}{2(t^2+t+1)}\ dt=\int_0^1  \dfrac{\log (t)}{2(t^2+t+1)}\ dt-\int_0^1  \dfrac{\log (t)}{2(t^2+t+1)}\ dt=0$
$\displaystyle \int_0^\infty  \dfrac{\log 2}{2(t^2+t+1)}dx=\left[\ \dfrac{\log 2}{\sqrt{3}}\arctan\left(\dfrac{1+2x}{\sqrt{3}}\right) \right]_0^\infty=\dfrac{\pi\log 2}{3\sqrt{3}}$
A: Well, I will present another method:

*

*Compute:

$$\mathscr{S}_\text{n}:=\frac{\partial}{\partial\text{n}}\left\{\int_0^\infty\frac{\ln\left(\text{n}x\right)}{x^2+2x+4}\space\text{d}x\right\}=\int_0^\infty\frac{1}{x^2+2x+4}\cdot\frac{\partial}{\partial\text{n}}\left(\ln\left(\text{n}x\right)\right)\space\text{d}x=$$
$$\int_0^\infty\frac{1}{x^2+2x+4}\cdot\frac{\partial}{\partial\text{n}}\left(\ln\left(\text{n}\right)+\ln\left(x\right)\right)\space\text{d}x=$$
$$\int_0^\infty\frac{1}{x^2+2x+4}\cdot\left\{\frac{\text{d}}{\text{dn}}\left(\ln\left(\text{n}\right)\right)+\ln\left(x\right)\cdot\frac{\partial}{\partial\text{n}}\left(1\right)\right\}\space\text{d}x\tag1$$


*Now, find:

$$\mathcal{I}:=\int_1^2\mathscr{S}_\text{n}\space\text{dn}\tag2$$


*Prove that the integral you want to find is equal to $\mathcal{I}$;

*You will/must find:

$$\mathcal{I}=\frac{\pi\ln\left(2\right)}{3\sqrt{3}}\tag3$$
