Integrate $\int_0^\infty \frac{\log x}{x^2+2x+4}\ dx$

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

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!

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,

$\displaystyle \int_0^\infty \dfrac{\log (t)}{2(t^2+t+1)}\ dt=\int_0^1 \dfrac{\log (t)}{2(t^2+t+1)}\ dt-\int_0^1 \dfrac{\log (t)}{2(t^2+t+1)}\ dt=0$

$\displaystyle \int_0^\infty \dfrac{\log 2}{2(t^2+t+1)}dx=\left[\ \dfrac{\log 2}{\sqrt{3}}\arctan\left(\dfrac{1+2x}{\sqrt{3}}\right) \right]_0^\infty=\dfrac{\pi\log 2}{3\sqrt{3}}$