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Math Help - INtegrals logarithm

  1. #1
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    INtegrals logarithm

    kindly check this integral:
    \int \frac{(\log_{3}x)^2}{x} dx

    let u = \log_{3}x
    du = \frac{\log_{3}e}{x}dx

    =\frac{1}{\log_{3}e}\int u^2 du

    =\frac{\ln{3}}{\ln{e}} \frac{u^3}{3}

    =\ln{3} \frac{(\log_{3}x)^3}{3} + C

    =\frac{\ln{3}}{3} (\frac{\ln{x}}{\ln{3}})^3 + C

    =\frac{\ln^3{x}}{3\ln^2{3}} + C
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    kindly check this integral:
    \int \frac{(\log_{3}x)^2}{x} dx

    let u = \log_{3}x
    du = \frac{\log_{3}e}{x}dx

    =\frac{1}{\log_{3}e}\int u^2 du

    =\frac{\ln{3}}{\ln{e}} \frac{u^3}{3}

    =\ln{3} \frac{(\log_{3}x)^3}{3} + C

    =\frac{\ln{3}}{3} (\frac{\ln{x}}{\ln{3}})^3 + C

    =\frac{\ln^3{x}}{3\ln^2{3}} + C
    If:

    <br />
u=\log_3(x)<br />
,

    then

    <br />
u=\frac{\ln(x)}{\ln(3)}<br />

    so:

    <br />
\frac{du}{dx}=\frac{1}{x \ln(3)}<br />

    RonL
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    kindly check this integral:
    \int \frac{(\log_{3}x)^2}{x} dx

    let u = \log_{3}x
    du = \frac{\log_{3}e}{x}dx

    =\frac{1}{\log_{3}e}\int u^2 du

    =\frac{\ln{3}}{\ln{e}} \frac{u^3}{3}

    =\ln{3} \frac{(\log_{3}x)^3}{3} + C

    =\frac{\ln{3}}{3} (\frac{\ln{x}}{\ln{3}})^3 + C

    =\frac{\ln^3{x}}{3\ln^2{3}} + C
    Checked. Absolutely no problem. My method leads to the same result.
    Attached Thumbnails Attached Thumbnails INtegrals logarithm-june30.gif  
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