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

6 years ago

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

6 years ago

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

6 years ago

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

6 years ago

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

6 years ago
OpenStudy (anonymous):

Apply U sub, it works fine!

6 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!!

6 years ago

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

6 years ago

but okay.. hmm...

6 years ago
OpenStudy (ash2326):

Yeah, did you get it?

6 years ago

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

6 years ago
OpenStudy (ash2326):

Correct:D

6 years ago

$\frac{t^4}{4}$ + c

6 years ago

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

6 years ago
OpenStudy (ash2326):

good work:D

6 years ago