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Thread: 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.

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

    I made my $\displaystyle u = log_3x$ and took the derivative.

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

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

    So now I have

    $\displaystyle \int 8ln(3)^2udu$

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

    $\displaystyle 4ln(3)^2(log_3x)^2+c$

    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.

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

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

    $\displaystyle \int 8ln(3)^2udu$
    Here we should explain. Actually, the notation $\displaystyle ln(3)^2=(ln(3))^2$. You should have a squared "ln" in the numerator because you multiply by $\displaystyle \frac{ln(3)}{ln(3)}$ and you have $\displaystyle ln(3)$ in the numerator. so, you should have a squared "ln(3)", What did you write here is the right notation for $\displaystyle ( ln(3) )^2$, 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:

    $\displaystyle 4ln(3)^2(log_3x)^2+c$

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

    But you can solve it easily. Note that $\displaystyle log_{3}(x)=\frac {ln(x)}{ln(3)}$
    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:
    $\displaystyle 8\int \frac{ln(x)}{x} dx$.
    Which can be solved by the simple substitution $\displaystyle u=ln(x)$.
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