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Math Help - Help with an indefinite integral.

  1. #1
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    Help with an indefinite integral.

    I got an answer for this problem, but I'm just wondering I did it right.

    \int \frac{8ln(3)log_3 x}{x}

    I made my u = log_3x and took the derivative.

    dx = \frac{1}{xln(3)du}

    I moved the ln(3) to the other side because I only have a 1/x in the original equation.

    So now I have

    \int 8ln(3)^2udu

    So I integrate and re-substitute and I wind up with:

    4ln(3)^2(log_3x)^2+c

    Did I do my math correctly?
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  2. #2
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    Quote Originally Posted by xwanderingpoetx View Post
    I got an answer for this problem, but I'm just wondering I did it right.

    \int \frac{8ln(3)log_3 x}{x}
    Where is the "dx" ?
    I made my u = log_3x and took the derivative.

    dx = \frac{1}{xln(3)du}
    I moved the ln(3) to the other side because I only have a 1/x in the original equation.
    I did not follow you here, You should multiply the integrated function by \frac{ln(3)}{ln(3)} ; to get the ln(3) in the denominator.
    So now I have

    \int 8ln(3)^2udu
    Here we should explain. Actually, the notation ln(3)^2=(ln(3))^2. You should have a squared "ln" in the numerator because you multiply by \frac{ln(3)}{ln(3)} and you have ln(3) in the numerator. so, you should have a squared "ln(3)", What did you write here is the right notation for ( ln(3) )^2, But clearly, you did not mean that; because you wrtie it as "ln(3)^2".
    So I integrate and re-substitute and I wind up with:

    4ln(3)^2(log_3x)^2+c

    Did I do my math correctly?
    hmmmm, seems correct answer for me.

    But you can solve it easily. Note that log_{3}(x)=\frac {ln(x)}{ln(3)}
    If you substitute this in the original integral and the two "ln(3)" will cancel.
    and take the "8" before the integral, You will face the following integral:
    8\int \frac{ln(x)}{x} dx.
    Which can be solved by the simple substitution u=ln(x).
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