Question

U-Substitution: Evaluate

integral of (20x^4/(4x^5 + 3))dx; u=4x^5 + 3

I found dx but I'm not sure if the answer is integral of (1/(4x^5 + 3))

Answer 1

\[u = 4x^5+3\] then we have \[\frac{ du }{ dx } = 20x^4 \implies du = 20x^4dx\] now compare it with your integrand, plug it all in.

Answer 2

yes, then dx = du/20x^4 and you would have
integral of (20x^4/u)(du/20x^4)
do i cancel out both 20x^4?

Answer 3

No, we don't solve for dx when we make the substitution, we are solving for du, since we want to simplify the integrand. Compare the integral with the substitution \[\int\limits \frac{ \color{red}{20x^4dx} }{ 4x^5+3 }\]

Answer 4

We replace everything in the integral in terms of u

Answer 5

okay, always solve for du and not dx?

Answer 6

Yes, if you're given a integral as such.

Answer 7

The denominator becomes u

Answer 8

Right

Answer 9

How about the numerator what happens

Answer 10

you find the integral of the numerator?
4x^5?

Answer 11

Ok lets take this slow, so we have our integral \[\int\limits \frac{ 20x^4 }{ 4x^5+3 }dx\] we start by noticing u - substitution is the best method as the degree in the denominator is a degree higher so it will work nicely. Then we make our u substitution \[u=4x^5+3\]\[\frac{ du }{ dx } = 20x^4 \implies du = \color{red}{20x^4dx}\] now we look back at our integral and compare, \[\int\limits \frac{ \color{red}{20x^4} }{ 4x^5+3 }\color{red}{dx}\] notice our du is exactly the same as when we did the derivative for du/dx. So plugging it all in since we are making it in terms of u we get, \[\int\limits \frac{ \color{blue}{du}}{ u }\] notice our integral has been simplified now since all we have to do is make our "substitution in terms of u."

Answer 12

okay, so its du/4x^5 + 3?

Answer 13

What do you mean

Answer 14

\[u=4x^5+3\] So we have \[\int\limits \frac{ du }{ u }\]

Answer 15

u = 4x^5 + 3 and if you plug that into du/u is it du/4x^5 + 3?

Answer 16

Nooo, read what I've said, we make the substitution in terms of u, so it's easier to integrate, we will plug our 4x^5+3 after the integration.

Answer 17

Now we just go ahead and integrate \[\int\limits \frac{ du }{ u }\]

Answer 18

okay, how do you integrate that?

Answer 19

Note that \[\int\limits \frac{ 1 }{ x } dx = \ln(x)+C\]

Answer 20

I figured ln had something to do with it. What is next?

Answer 21

Please try it yourself, see what you get

Answer 22

ln(4x^5 + 3)

Answer 23

Exactly, don't forget the +C :)

Answer 24

reeeeaaaaaallly?! wow, thank you! Can you help me with another please?

Answer 25

Sure, I'll try

Answer 26

Same directions

\[\int\limits 36x^2e^(4x^3 + 3) dx\]

The (4x^3 + 3) is raised

Answer 27

Cool, so any idea where to begin? I will let you do most of it this time as it's good practice.

Answer 28

u = 4x^3 + 3

du = 12x^2dx

Answer 29

Good, what's next

Answer 30

(36x^2e^u)dx

Answer 31

Mhm, \[\int\limits 36x^2e^{4x^3+3}dx\]
\[u = 4x^3+3 \implies du = 12x^2dx \implies \frac{ du }{ 12 }=x^2dx\] again compare this with your original integral and make all the substitutions correctly.

Answer 32

why do you divide the 12?

Answer 33

Is there a 12 in the integrand originally?

Answer 34

We are replacing what already exists

Answer 35

so because there was no 12 in the original but a x^2, you have to divide by the 12 so the x^2 is alone?

Answer 36

Right, because we are "substituting" in for what already exists, if you remember high school math when we had substitution for systems of equations you were replacing equations with substitution.

Answer 37

A lot of the math goes back to what you learnt before, I always say the calculus part is easy it's the math you've learnt before that you need to remember.

Answer 38

okay. so is 36x^2e^(u) correct or is 36(u)e^(4x^5 + 3) correct since we already have x^2?

Answer 39

I know, I understand most parts of calculus but i always have trouble remembering the algebra

Answer 40

It seems you're still having a hard time grasping what substitution means, we don't want the integrand to be a product, that would make it problematic for you. We don't want something like x*e^x this is bad for now, but your integral is set up that requires a "substitution" which gets rid of the product and makes it easy to integrate.

Answer 41

so what do i have to do after I find that x^2dx = du/12?