Convergence of $\int_{2}^{\infty }\frac{1}{x*\ln ^2x}dx $ equals $\frac{1}{\ln 2}$. Why?

Question

Given :$$\int_{2}^{\infty }\frac{1}{x\ln ^2x}dx $$

marking 'lnx' as 't' I get: $$\int_{2}^{\infty }\frac{1}{e^tt^2}dt $$

Doing integration by parts I get: $$-\frac{1}{t^2e^t}+2\int te^t dt$$

Repeating the process of integration by parts: $$-\frac{1}{t^2e^t}+2te^t-\int e^tdt$$ which equals to: $$-\frac{1}{t^2e^t}+2te^t-e^t$$

putting back the X's: $$-\frac{1}{x{(\ln x)}^2}+2x\ln x-x$$

Is the integration valid? Because after placing x as infinity and 2 respectively I get a divergent integral, contrary to the answer which is 1/ln2

Answer 1

Hint: Substitute $$t=\ln(x)$$ then we get $$dt=\frac{1}{x}dx$$

Answer 2

$$\ln x =t$$ $$x=e^t$$ $$dx=e^t dt$$ $$dx=xdt$$ $$I=\int_{2}^{\infty }\frac{1}{x\ln ^2x}dx$$ $$I=\int_{\ln 2}^{\infty} \frac{dt}{t^2}$$