Question

Hey all,
I need some calc help. Iv'e done most but still need help. Its the indefinite integral or antiderivative. THNXS!

Answer 1

We looking at number 2 or 3? :)

Answer 2

Oh also,
\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 3

both because I keep getting a different answer from the answer choices

Answer 4

\[\Large\bf\sf \int\limits \frac{x^2}{(2x^3-12)^4}\;dx\quad=\quad \int\limits \frac{1}{(\color{royalblue}{2x^3-12})^4}\left(\color{orangered}{x^2\;dx}\right)\]We want to make the substitution,\[\Large\bf\sf \color{royalblue}{u=2x^3-12}\]

Answer 5

We'll take the derivative of our substitution, and we're hoping that it gives us something close \(\Large\bf\sf \color{orangered}{\text{to this}}\).

Answer 6

Understand how to take the derivative of u? :)

Answer 7

ok one sec

Answer 8

then what?

Answer 9

Well if you calculated your derivative correctly, you should get,\[\Large\bf\sf du=6\color{orangered}{x^2\;dx}\]
See how it `almost` matches the orange thing in our integral?

Answer 10

yup

Answer 11

Hmm so to make them match, we'll have to get rid of this 6.

Answer 12

Dividing both sides by 6 should do the trick.

Answer 13

\[\Large\bf\sf \color{orangered}{\frac{1}{6}du=x^2\;dx}\]

Answer 14

\[\Large\bf\sf \int\limits\limits \frac{1}{(\color{royalblue}{2x^3-12})^4}\left(\color{orangered}{x^2\;dx}\right)\quad=\quad \int\limits\limits \frac{1}{(\color{royalblue}{u})^4}\left(\color{orangered}{\frac{1}{6}\;du}\right)\]

Answer 15

Understand how I plugged those in? :o

Answer 16

yup

Answer 17

Let's pull the 1/6 outside of the integral since it's constant.
Then we'll apply a handy rule of exponents to make this easier to deal with.\[\Large\bf\sf =\quad\frac{1}{6}\int\limits u^{-4}\;du\]

Answer 18

From here we can apply the `Power Rule for Integration`.

Answer 19

Remember how to do that? :)
Do it!!!

Answer 20

could u review me plz

Answer 21

Noooo, put in some effort little lady! :) lol

I'll show you an example:\[\Large\bf\sf \int\limits x^n\;dx \quad=\quad \frac{1}{n+1}x^{n+1}\]It's two steps:
`Add 1 to the exponent`
`Divide by the new exponent`

Example:\[\Large\bf\sf \int\limits x^{-11}\;dx\quad=\quad \frac{1}{-11+1}x^{-11+1}\quad=\quad -\frac{1}{10}x^{-10}\]

Answer 22

oh is that the power rule?

Answer 23

Yes. It's like the Power Rule for Derivatives, but in reverse, and in the reverse order as well.

Answer 24

ok give me a few min so I can try it and I'll ask you if my answer is right k?

Answer 25

k c:

Answer 26

(1/6)(-1/3)u^-3, @ganeshie8 can u check my work

Answer 27

@

Answer 28

2

Answer 29

\(\large \color{red}{\checkmark}\)

Answer 30

u just need to change it to x's again

Answer 31

\(u\)'s are not there in the given integral
you have introduced it
so you need to replace it back wid \(x\)'s

Answer 32

i see thnxs

Answer 33

what about the plus C

Answer 34

\((1/6)(-1/3)u^{-3} + c\)

\(-1/18u^{-3} + c\)

\(-1/(18u^3) + c \)

Answer 35

just put it, c always comes wid the indefinite integral

Answer 36

thank you so much!

Answer 37

\(-1/(18u^3) + c\)

\(\large \frac{-1}{18(2x^3-12)^3} + c\)

Answer 38

np :) hope u can try the next one...

Answer 39

I can do all of them now thnxs

Answer 40

just to give u a headstart :-
for question #3,
substitute \(u = 2 + \sqrt{x}\)

Answer 41

Good :)

Answer 42

@zepdrix I figured it out thank you so much!

Answer 43

Oh cool c: