Question

evaluate the integral
18t^(1/3)ln(t)

Answer 1

u=lnx and dv=18t^(1/3)

Answer 2

yes dv = 18t^1/3)dt

Answer 3

so u = d(ln(x))/dt and v = int(dv)
u = 1/x and v = 9/2*x^4

Answer 4

okay any need to figure out what du is?

Answer 5

sorry du = 1/t dt

Answer 6

replace any x with t..sorry

Answer 7

no worries so how would we put that into the formula?

Answer 8

\[uv - \int\limits_{}^{} v * du\]

Answer 9

u = ln(t) v = (9/2) *t^4 du = (1/t) dt

Answer 10

\[(\ln(t)*(9/2)t^4 - \int\limits_{}^{} (9/2) * x^4 * (1/x) dt\]

Answer 11

\[\ln(t)*(9/2)t^4- (9/2)\int\limits_{}^{}(t^4/t) = \ln(t)*(9/2)t^4- (9/2)\int\limits_{}^{}(t^3) dt\]

Answer 12

\[\ln(t)*(9/2)t^4- (9/2)(1/4)t^4 = \ln(t)*(9/2)t^4- (9/8)t^4 \]

Answer 13

would that be ln(t)*-(9/6)t^4?

Answer 14

oopes i meant ln(t)*(-9/6)t^4

Answer 15

or is that the wrong way of structuring it

Answer 16

sec phone

Answer 17

ok :)

Answer 18

i would factor out (9/2) t^4
and get (9/2)*t^4(ln(t)-(1/4))

Answer 19

\[(9/2)*t^4*(\ln(t)-(1/4))\]

Answer 20

wait this is wrong I forgot about the 18 sec sorry im doing lots of stuff

Answer 21

hmm homeworks saying it isnt correct maybe..

Answer 22

oh okay take your time!!

Answer 23

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

Answer 24

so (27/2)*t^(4/3)*ln(t) - (27/2)int(t^(1/3)dt

Answer 25

= (27/2)*t^(4/3)*ln(t) - (81*t^(4/3))/8 + c

Answer 26

(27/8) t^(4/3)(4ln(t)-3)+c

Answer 27

could that be simplified anymore?

Answer 28

depends on what you call simple:)

Answer 29

sec let me see if wolfram wil do this...
http://www.wolframalpha.com/input/?i=integrate+18t%5E%281%2F3%29ln%28t%29
this is a great place to look at some integrals...notice how I just typed in what I wanted at the top

Answer 30

hit the show steps button.

Answer 31

okay ill try that. i was just asking because my hw was saying that the answer you have me was incorrect :/

Answer 32

i found a form that worked thank you so much!! that site is very helpful!