How to Integrate $\frac{1}{x(lnx)^2}$ Using $u$-Substitution

[karush]

$\tiny{242.2q.3a}$
\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}
$\textit{think this ok, wasn't sure on simplifying}$

 

[MarkFL]

We are given:

\(\displaystyle I=\int_2^4 \frac{1}{x\ln^2(x)}\,dx\)

After making your $u$-substitution, we have:

\(\displaystyle I=\int_{\ln(2)}^{2\ln(2)} u^{-2}\,du=-\left[u^{-1}\right]_{\ln(2)}^{2\ln(2)}=\frac{1}{\ln(2)}-\frac{1}{2\ln(2)}=\frac{1}{2\ln(2)}\)

You did fine, although when making the substitution I would change the limits in accordance with the substitution, and then simplify the final answer. :D