Question

Evaluate the integral using integration by parts where possible.
t^(1/3)lnt dt

Answer 1

\[u=\ln t\]\[dv=t^{1/3}\]\[\int udv=uv-\intvdu\]try it

Answer 2

*\[\int udv=uv-\int vdu\]try it

Answer 3

dt = 1/t du right?

Answer 4

du=(1/t)dt

Answer 5

and v = (t^4/3)/(4/3)?

Answer 6

yes

Answer 7

Okay, I got to (ln t)((t^4/3)/(4/3)) - I((t^4/3)/(4/3))*t^1/3
Where I = Integral.
Now I'm lost. The Integral of ((t^4/3)/(4/3)) would be fraction on top of fraction..is that supposed to happen?

Answer 8

gotta remember a little algebra sometimes ;)\[\frac1{4/3}=3/4\]so\[\int t^{1/3}\ln t=\frac34t\ln t^{4/3}-\frac34\int t^{4/3}\frac1tdt\]