Let $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$ Then $$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$ which is the integral in question.

On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence the integral in question is $$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2 + \Gamma''(1)/2 =\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where $\gamma=0.57721\dots$ is the Euler gamma constant.

Let $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$ Then $$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$ which is the integral in question.

On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence the integral in question is $$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2 + \Gamma''(1)/2 =\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where $\gamma=0.57721\dots$ is the Euler gamma constant.

(The second equality in the latter display follows because $\Gamma'(1)=-\gamma$ and $\Gamma''(1)=\gamma^2+\pi^2/6$. In turn, the latter two equalities can be obtained using the last two displays in Section Recurrence relation.)

The integral is $$f''(1)=\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where $\gamma=0.57721\dots$ is the Euler gamma constant, and $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx=2^{-a}\,\Gamma(a).$$

Let $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$ Then $$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$ which is the integral in question.

On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence the integral in question is $$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2 + \Gamma''(1)/2 =\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where $\gamma=0.57721\dots$ is the Euler gamma constant.