Techniques of estimating integrals, asymptotic behavior of Li(x)

Question

I need help understanding the logic and build an rigorous intuition proving that if $x \gt 2$, then:

$$\int_{2}^{x} \frac{dt}{t \log^{3} t} \lt \frac{1}{2 \log^{2}2}$$

Answer 1

Notice that $$\int_2^x \frac{dt}{t\ln^3(t)}<\int_2^{\infty} \frac{dt}{t\ln^3(t)}=\frac{1}{2\ln^2(2)}$$ and we are done.

Answer 2

$$\int_{2}^{x} \frac{dt}{t \log^{3} t} = \frac12\left(\frac{1}{\log^{2}2}-\frac{1}{\log^{2}x}\right)<\frac{1}{2\log^{2}2}$$ How much logic would one need for that? How much logic can one muster while seriously combining "rigorous" and "intuition"?