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Math Help - u-substitution problem help

  1. #1
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    u-substitution problem help

    I need help with a u-substitution problem. It is has already been worked out for me but I don't understand how to get the final answer.
    Here is the problem:

    ∫ln(x)/x

    here is the worked out answer:

    u=ln(x) ---> ∫udu=(u^2)/2 + C
    =(ln(x))^2)/2 + C

    I don't understand why ln(x) changes to u^2 when you substitute u into the integral, and I also don't understand where the 2 on the bottom comes from. I thought the substitution should be u/x if you substitute u for ln(x) into the integral. Maybe a step was skipped in the solution and that's why I don't understand? Anyway, if anyone could explain to me why it becomes u^2/2 + C instead of u/x, I would greatly appreciate it. Thanks in advance to anyone who can help.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Matt164 View Post
    I need help with a u-substitution problem. It is has already been worked out for me but I don't understand how to get the final answer.
    Here is the problem:

    ∫ln(x)/x

    here is the worked out answer:

    u=ln(x) ---> ∫udu=(u^2)/2 + C
    =(ln(x))^2)/2 + C

    I don't understand why ln(x) changes to u^2 when you substitute u into the integral, and I also don't understand where the 2 on the bottom comes from. I thought the substitution should be u/x if you substitute u for ln(x) into the integral. Maybe a step was skipped in the solution and that's why I don't understand? Anyway, if anyone could explain to me why it becomes u^2/2 + C instead of u/x, I would greatly appreciate it. Thanks in advance to anyone who can help.
    When you make a substitution, you are changing the integral into something you can easily integrate. In integrating \int\frac{\ln(x)}{x}\,dx, you make the substitution {\color{blue}u=\ln(x)}. Differentiating, we get \frac{\,du}{\,dx}=\frac{1}{x}\implies {\color{red}\,du=\frac{\,dx}{x}}

    Now let's rexamine the integral:

    \int\frac{{\color{blue}\ln(x)}}{{\color{red}x}}{\c  olor{red}\,dx}

    Making the proper substitutions, we end up with \int u\,du

    Integrating, we end up with \tfrac{1}{2}u^2+C (we need a constant since this is indefinite integration [no limits])

    But we were after \int\frac{\ln(x)}{x}\,dx, so now backsubstitute u=\ln(x) back into your antiderivate to now get \color{red}\boxed{\tfrac{1}{2}\left(\ln(x)\right)^  2+C}

    Does this process make a little more sense now?
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