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  1. MHB Master
    karush's Avatar
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    #1
    $\tiny{242.2q.3a}$
    \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}
    $\textit{think this ok, wasn't sure on simplifying😎}$

  2. Pessimist Singularitarian
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    #2
    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.

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