Question

I can't intergrate this for some reason lol...

Answer 1

im doing integration right now too. I can look and see if i can help but I am learning too

Answer 2

\[\int\limits x^2 (x^3 -1)^4\]

Answer 3

dx*

Answer 4

btw with u substitution

Answer 5

i thing u = x^3 -1

Answer 6

brb

Answer 7

you could expand the quartic equation then multiply by x^2

or the better choice is to use a substitution

\[u = x^3 - 1\]

Answer 8

that's a great choice so what is

\[\frac{du}{dx}= ?\]

Answer 9

x^4

Answer 10

but there is no x^4 only x^2 thats where i cant get my head around it

Answer 11

no look at your substitution and find the derivative of u with respect to x

Answer 12

sry its 3x^2

Answer 13

wasn't thinking i have that written 100 times lol i am struggling

Answer 14

if you use your substitution of

\[u = x^3 - 1\]

then

\[\frac{du}{dx} = 3x^2\]

does this help?

Answer 15

no i knew that I just messed up typing it here that is where i am stuck... the book only gives the answer

Answer 16

do i bring the 3 out in front

Answer 17

oops should be\[\int\limits (x^3 -1)^4 x^2 dx = \int\limits u^4 \times \frac{1}{3} du\]

Answer 18

close you take 1/3 outside

\[\frac{1}{3} \int\limits u^4 du\]

Answer 19

okay awesome... big time help, thank you