Question

Calc II (Single Var) || Substitution Rule (u-sub integrals) Review... LaTeX below

Answer 1

\[\int4x(2x^2+3)^7\]

Answer 2

let u=something
when you take the derivative of *something * you should be able to cancel out the variable.

Answer 3

u = 4x ?

Answer 4

No, you must make u=2x^2+3

Answer 5

du = 4x dx

Answer 6

If I do that, I don't get a dx

Answer 7

yes you do

Answer 8

hmm well remember the f o g composition function frm algebra classes right ?
(2x^2+3) raised to the 7 power

Answer 9

well if I put that as u then I get du=2x
what am I supposed to do from there o_O

Answer 10

well if you let u=4x
how would the derivative of u help you to cancel out ?

Answer 11

du = 4dx?
idk but I got stuck on step one if I make u = 2x^2 + 3

Answer 12

\[\int\limits_{ }^{} 4x(2x^2+3)^7 \color{Red}{dx} \]
let u =2x^2+3
\ solve that equation for `dx` \[dx=\frac{du}{4x}\] now we can replace dx with the fracation du/4x \[\int\limits_{ }^{} 4x(2x^2+3)^7 \cdot \color{Red}{\frac{du}{4x}}\]

Answer 13

it doesn't work when I write latex :(

how did dx become that o_O

Answer 14

\[\large\rm \int\limits_{ }^{} 4x(\color{green}{u})^7 \cdot \color{Red}{ \frac{du}{4x} }\]
**

Answer 15

https://www.symbolab.com/solver/step-by-step/%5Cint4x%5Cleft(2x%5E%7B2%7D%2B3%5Cright)%5E%7B7%7Ddx

it put u = x^2 I'm confused

Answer 16

This

Answer 17

because you're taking the derivative of u with respect to x
\[\large\rm u=2x^2+3 \rightarrow \frac{du}{dx}=4x \frac{dx}{dx}\]

Answer 18

\[\frac{du}{dx}=4x\]
dx cancels each other out
so we are left with du/dx=4x
if you multiply both sides by dx
you will get ` du =4x dx `

Answer 19

You can do that as well
but i find u=2x^2+3 easy
the whole point of u-substitution is to cancel out the x variable

Answer 20

is it ok to take out the 4 like it says here?

Answer 21

sorry there is major lag and loading issues over here... I might be missing the obvious so apologies

Answer 22

That's totally fine you can take constant \[\int\limits_{ }^{ } c~ dx \rightarrow c \int\limits_{ }^{} 1~ dx\]

Answer 23

out**

Answer 24

alright then thank you
I think I get the idea from there on out :)

Answer 25

you're welcome