# Math Help - INtegrals logarithm

1. ## INtegrals logarithm

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

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

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

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

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

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

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

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:

$
u=\log_3(x)
$
,

then

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

so:

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

RonL

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.