# Thread: INtegrals logarithm

1. ## INtegrals logarithm

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

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

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

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

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

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

$\displaystyle =\frac{\ln^3{x}}{3\ln^2{3}} + C$

2. Originally Posted by ^_^Engineer_Adam^_^
kindly check this integral:
$\displaystyle \int \frac{(\log_{3}x)^2}{x} dx$

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

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

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

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

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

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

$\displaystyle u=\log_3(x)$,

then

$\displaystyle u=\frac{\ln(x)}{\ln(3)}$

so:

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

RonL

3. Originally Posted by ^_^Engineer_Adam^_^
kindly check this integral:
$\displaystyle \int \frac{(\log_{3}x)^2}{x} dx$

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

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

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

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

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

$\displaystyle =\frac{\ln^3{x}}{3\ln^2{3}} + C$
Checked. Absolutely no problem. My method leads to the same result.