Question

can anyone help me to integrate this respect to x; (x-1)^3. i've forgotten how to do it.

Answer 1

u know this basic integration formula :
\(\huge \int x^ndx=?\)

Answer 2

i know that, but this one has power. do we separate the x and 1 and integrate them separately?

Answer 3

it will follow the same formula as you would do for x^3

Answer 4

3(x-1)^2 ?

Answer 5

@anybody integration not differentiation

Answer 6

will somebody explain why we need a u-sub please, I think that is where the asker is having difficulty

Answer 7

\[\frac{ (x-1)^{3} }{ 4 }\]?

Answer 8

\[\int(x-1)^3dx\]now notice that dx is the differential of the argument x-1, so we call it u:
\[u=x-1\implies du=dx\]after the sub the integral is then\[\int u^3du\]can you integrate that?

Answer 9

u^4/4

Answer 10

great, now sub back in for what u is...

Answer 11

oh and always include the +C if it is indefinite

Answer 12

where do we get the +c?

Answer 13

that's a different question, all indefinite integrals (integrals where the bounds are not specified) need a +C at the end where C is some unknown constant
if you want an explanation for that post it separately please
for now just sub in what we had fopr u into u^4/4 and add a +C at the end and tell me what you get

Answer 14

(x-1)^4/4

Answer 15

+C
;)

Answer 16

okey, thanks ;)

Answer 17

welcome
if you want me to explain the +C part I can try, but it is a little subtle, and please post it as a separate question
welcome!