Question

integrate (lnx)^3

Answer 1

Help me

Answer 2

\[\int (\ln x)^3\text d x=3\int\ln x \text d x=3(x\ln x-x)+c\]

Answer 3

oh no no
\[\ln(x^3)=3\ln(x) \text{ but } (\ln(x))^3 \neq 3 \ln(x)\]

Answer 4

oh yeah unkle you're wrong

Answer 5

this is a by parts question

Answer 6

dx = du

v=ln(x)

Answer 7

solving using by parts and the ^

Answer 8

this is best done with the IBP probably

u = (ln x)^3
du = 3 (ln x)/x

dv = dx
v = x

\[\large x (\ln x)^3 - \int 3 \ln x dx\]

Answer 9

did he do by parts, i can't even tell and since you're already answering i feel no need to actually solve it

Answer 10

i assume you can do it from there @isham_92 ?

Answer 11

is du right?

Answer 12

or did he think it was ln(x^3)

Answer 13

yeah dx=du

Answer 14

so you get x=u

Answer 15

wahhh wrong >.<

Answer 16

do i have to actually solve it -.-

Answer 17

...dont make me please lol

Answer 18

u = (ln x)^3
du = 3(ln x)^2/x

dv = dx
v = x

\[x (\ln x)^3 - \int 3(\ln x)^2 dx\]

this needs another integration by part :/

Answer 19

yessirrr

Answer 20

until you get to ln(x)^1

Answer 21

i do believe this is on the integration table tho not sure tho =/

Answer 22

\[3\int (\ln x)^2 dx\] u = (ln x)^2
du = (2 ln x) /x

dv = dx
v = x

\[x (\ln x)^2 - \int 2 \ln x dx\]
\[x(\ln x)^2 - 2x \ln x +2x\] so inserting in the original one... \[x (\ln x)^3 - 3[x (\ln x)^2 - 2x \ln x + 2x]\]
\[x (\ln x)^3 - 3x (\ln x)^2 + 6x \ln x - 6x\]

that right @myininaya

Answer 23

i was way off, sorry about that