Question

@perl

Answer 1

@perl help plz!!

Answer 2

can you show me any work you did so far

Answer 3

so i can distribute first right?

Answer 4

i havnt dtarted this problem because im confused

Answer 5

start with a u substitution
u = x^3 - 4

Answer 6

ohh ok i know how to do this problem, one second while i colve it out, will you check it for me after?

Answer 7

ok

Answer 8

@perl i got 1/4(x^3-4)^4+C , could you please check? @perl

Answer 9

i think i did it correctly, im trying to upload my work because i did it on a piece of paper

Answer 10

when you do your u-sub you have another 1/3 overthere. Incorrect.

Answer 11

u=x^3-4

du=3x^2 dx ---> (1/3) du = x^2 dx

Answer 12

So
Integral of x^2 (x^3-4) dx

the x^3-4 gets replaced by u
the x^2 dx get replaced by (1/3) du

you have (1/3) Integral u^4 du

Answer 13

and when you integrate, you divide by 1/4 yes, but you ALREADY have the 1/3 there, so in the final answer you have a coefficient of 1/12 (not 1/4).

Answer 14

Is this making any sense /

Answer 15

??

Answer 16

ohh so my final answer would be 1/12(x^3-4)^4+C? @perl

Answer 17

* @SolomonZelman

Answer 18

yes, and do you fully understand why ?

Answer 19

yes!

Answer 20

very nice... have a good night, I am going offline:)

Answer 21

@perl

Answer 22

@perl ????