$\tiny{242.2q.3a}$

\begin{align}\displaystyle

I_{3a}&=\int_2^{4} \frac{1}{x(ln \, x)^2} \, dx\\

u&=\ln \, x \therefore du=\frac{1}{x}\, dx

\end{align}

$\textsf{substitute $\ln \, x =u$}$

\begin{align}\displaystyle

I_u&=\int_2^{4} \frac{1}{u^2} \, du\\

&=-\left[\frac{1}{u}\right]^{4}_2 \\

%&=\frac{1}{u} \right|_2-\frac{1}{u} \right|^{4}\\

\end{align}

$\textsf{back substitute $u=\ln \, x $}$

\begin{align}\displaystyle

&I_{3a}=\frac{1}{\ln \, 2 }

-\frac{1}{2\ln \, 2 }

\end{align}

$\textit{think this ok, wasn't sure on simplifying😎}$