OpenStudy (konradzuse):

integration by parts ((ln(x))^3)/x dx

5 years ago
OpenStudy (konradzuse):

u = ln(x) du = 1/x dx dv = x v = x^2 /x

5 years ago
OpenStudy (konradzuse):

\[\frac{x^2}{2} *\ln(x) -\frac{1}{x}\int\limits \frac{x^2}{2} dx\]

5 years ago
OpenStudy (konradzuse):

\[\frac{x^2}{2} * \ln(x) - \frac{1}{x} * \frac{x^3}{6} + c?\]

5 years ago
OpenStudy (konradzuse):

not sure if I did hat right... Idk if I am able to pull out that 1/x?

5 years ago
OpenStudy (anonymous):

Apply U sub, it works fine!

5 years ago
OpenStudy (ash2326):

@KonradZuse We don't need to do by parts here \[\int \frac{(\ln x)^3}{x}dx\] put \[\ln x= t\] we get \[\frac{dx}{x}=dt\] so our integral becomes \[\int t^3 dt\] I think you can solve from here!!

5 years ago
OpenStudy (konradzuse):

oh crap I messed up... it should have been x^-1

5 years ago
OpenStudy (konradzuse):

but okay.. hmm...

5 years ago
OpenStudy (ash2326):

Yeah, did you get it?

5 years ago
OpenStudy (konradzuse):

yeah makes sense. dx/x = dt or dx = xdt? so it would be xt^3/x so the x's cancel out.

5 years ago
OpenStudy (ash2326):

Correct:D

5 years ago
OpenStudy (konradzuse):

\[\frac{t^4}{4}\] + c

5 years ago
OpenStudy (konradzuse):

\[\frac{(\ln(x))^4}{4} + c\]

5 years ago
OpenStudy (ash2326):

good work:D

5 years ago
OpenStudy (konradzuse):

:)

5 years ago
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