Question

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

Answer 1

u = ln(x)
du = 1/x dx

dv = x

v = x^2 /x

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Apply U sub, it works fine!

Answer 6

@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!!

Answer 7

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

Answer 8

but okay.. hmm...

Answer 9

Yeah, did you get it?

Answer 10

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

Answer 11

Correct:D

Answer 12

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

Answer 13

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

Answer 14

good work:D

Answer 15

:)