Question

If I have the integral r^4lnrdr, should it be solved using parts?

Answer 1

\[ \int\limits_{1}^{3} r^4lnr dr\]

Answer 2

Ohhhh

Answer 3

Okay...

Answer 4

Yes....u = r^4 v' = ln r

Answer 5

Yes, U=r^4 dU=4r^3 v=x*ln(x)-x dv=ln(x)

Answer 6

Sorry, is with r xD... U=r^4 dU=4r^3 v=r*ln(r)-r dv=ln(r)

Answer 7

\[U=r^4 dU=4r^3 v=r*\ln(r)-r dv=\ln(r)\]

Answer 8

\[\int\limits r^4 \log (r) dr = 1/25*R^5 (5\log (r) -1) + c\]

Answer 9

For the above solution....you'd have to know the integral for ln(r), which is not a common integral. I prefer u=ln(r), dv=r^4dr....work from there.

Answer 10

Yeah, I goofed up, u is supposed to be ln(r)