Question

integral in lnx (x^-1/3) dx

Answer 1

Is this your integral?\[\int\frac{\ln x}{\sqrt[3]{x}}~dx\]

Answer 2

yes, it is

Answer 3

Integrate by parts. Let \(u=\ln x\) and \(dv=x^{-1/3}~dx\). It works out rather nicely.

Answer 4

that was what i do, but i'm stuck in ln \[\ln x (3/2 x^\frac{ 2 }{ 3 }) -\int\limits 3/2x^\frac{ 2 }{ 3 } (\frac{ dx }{ x }) \]

Answer 5

\[\frac{3}{2}x^{\frac{2}{3}}\ln x-\frac{3}{2}\int \dfrac{x^{\frac{2}{3}}}{x}~dx\]
Simplify: \(\dfrac{x^{\frac{2}{3}}}{x}=x^{\frac{2}{3}-1}=x^{-\frac{1}{3}}\)
\[\frac{3}{2}x^{\frac{2}{3}}\ln x-\frac{3}{2}\int x^{-\frac{1}{3}}~dx\]

Answer 6

thanks (y) i'll complete it

Answer 7

You're welcome!

Answer 8

\[\ln x -\frac{ 3x }{ 2 }^{\frac{ 2 }{ 3 }} -\frac{ 9x }{ 4 }^{\frac{ 2 }{ 3 }} + c\]

Answer 9

Where did the extra term come from? It should be
\[\frac{3}{2}x^{\frac{2}{3}}\ln x - \frac{9}{4}x^{\frac{2}{3}}+C\]

Answer 10

yeah, is't ok, thanks i'm a little wrong