Question

Find the indefinite integral

Answer 1

\[\int\limits_{?}^{?} 2\frac{ x^{2} }{ \sqrt[3]{x^{3}+2} } dx\]

Answer 2

I am assuming I use substitution.

Answer 3

i would let u = x^3 +2

Answer 4

can i take my 2 out?

Answer 5

then it follows

du = 3x^2dx

Answer 6

yes, definitely factor out the constant (2)

Answer 7

so du=3x^{2} dx?

Answer 8

sure does

Answer 9

what does your integral expression look like now

Answer 10

\[2\int\limits_{?}^{?}\frac{ x^{2} }{ u^\frac{ 1 }{ 3 } }\]

Answer 11

I took a little different path. note that du / 3 = x^2dx so I got:

\[2/3\int\limits_{}^{}u^(-2/3)\]

Answer 12

that should be u^(-2/3)

Answer 13

how does du/3=x^2 dx?

Answer 14

du = 3x^2dx --- Divide both sides by 3

du / 3 = x^2dx

Answer 15

why would i want to know du/3

Answer 16

because now you can divide your constant by 3. du = 3x^2dx has a constant in it which would just get in the way.

Answer 17

Keep your terms simple, and get rid of constants when you can

Answer 18

where does the 3 come from?

Answer 19

(basically why I got 2/3 on the outside)

Answer 20

oh i see where the 3 comes from.

Answer 21

the derivative of x^3 is 3x^2, that is where the 3 comes from initially.

Answer 22

so when i do these types of problems and there is a constant in my du value, i should divide by that constant?

Answer 23

yes, if you have something like du = (constant)x^3/sin^2(x)/cos(x)/etcdx

just divide the constant out so you get
du / constant = simplified-x-term-dx

Answer 24

\[2/3\int\limits_{?}^{?}\frac{ 2x }{ (x^{3}+2)^{2/3} }\]

Answer 25

so the du/3=x^2 dx goes in place of du then?

Answer 26

notice x^2dx = du so we can just substitute for that

so you have

2/3 (integral sign) u^(-2/3) du

u^(-2/3) is the same as 1 / u(2/3) --- I just moved the u up to the numerator.

Answer 27

1/ u^(2/3) ***

Answer 28

so the du/3=x^2 dx goes in place of du then?

Answer 29

like du/3 is the same thing as du now?

Answer 30

other way around, du goes in the place of the x^2

Answer 31

du/3 is the same as x^2dx which is what we had in the numerator, so we can just place the numerator with du

Answer 32

replace*

Answer 33

du / 3 is du because we already divided by 3 on the outside

Answer 34

im just a little confused as to where the 3 actually fits in the calculations. like, when i sub in du i place x^2 dx even though that's what we found for du/3

Answer 35

so because we divided by 3, when i take it outside the integrel is becomes 1/3

Answer 36

2/3 because you still have the first 2 you took out

Answer 37

you're basically multiplying it the outside by 1/3, you're just moving the du inside the integral now.

Answer 38

was the answer i wrote above correct?

Answer 39

no, you do not need 2x on the top. we only took the derivative of what we substituted originally which was x^3 +2

Answer 40

we need x^2 on the top so that we CAN substitute du for it. that's the whole point of integration by substitution

Answer 41

so its 2/3 int 1/(x^3+2)^2/3 x^2 dx?

Answer 42

before we substitute x^2 for du, the integral looks like

2/3 (integral sign) x^2 / (x^3 +2)^(2/3) dx

but we know x^2dx = du

so substitute du for x^2 and u for (x^3+2)

2 / 3 (integral sign) du / (u)^(2/3

= 2/3 (integral sign) u^(-2/3)du

Answer 43

still confused?

Answer 44

yup thats what i got.

Answer 45

good job!

Answer 46

can you solve from there?

Answer 47

well shouldnt it be to the power of 1/3 not 2/3?

Answer 48

since it was the cubed root

Answer 49

-1/3

you're right, it wouldnt be to the -2/3

Answer 50

(negative because im speaking in terms of moving everything up to the numerator. it would be positive in the denominator)

Answer 51

when you have

2/ 3 (Integral sign) du / CubedRoot[u] and you evaluate the integral, then it becomes

u^(2/3) + C

that's where I was confusing myself.

Answer 52

Then just substitute back for u, and you should be good!

Answer 53

got it! would you be able to help me with another?

Answer 54

I've got about 20 minutes. let's see how much we can get done!

Answer 55

int 3x cos(1-x^2) dx.

i got u=cos(1-x^2) and du=2xsin(x^2) dx

Answer 56

I would let u = 1-x^2

when you use substitution, generally don't include things like cos/ln/log/sin etc

Answer 57

okay i find this one confusing due to the 3x in there

Answer 58

Also, factor out that 3 so you get

3 (integral sign) xcos(1-x^2)

Answer 59

okay then i found du=2x dx

Answer 60

you should get

du = -2xdx

Answer 61

thats what i meant

Answer 62

if you were finding the derivative of 1 + x^2, then it would be du = 2xdx, but since it is minus, the derivative is

du = -2xdx

Answer 63

yah i know thats what i meant to write/

Answer 64

okay, then apply the same trick as last time

-du/2 = xdx

Answer 65

what does the integral become now?

Answer 66

3 int x cos(1-x^2) x dx?

Answer 67

-3/2 on the outside

-3/2 (integral sign) cos(u) du

remember we found du = xdx, so just replace the x with du

the -1/2 goes to the outside.

Answer 68

all that came from -du / 2 = xdx

and the cos(u) came from substituting u, with (1-x^2)

Answer 69

substituting (1-x^2) with u rather

Answer 70

i get confused at the du/2 part when i sub it back into the equation .. so now i replace the x that was in front of cos with du?

Answer 71

yup, that's exactly what we did when we had x^2dx = du /3 in the last equation.

when you have something like du / x

ALWAYS put x under what you have already on the outside (in this case it's 3) Also because it was - du / 2 put that NEGATIVE sign outside as well (which is why we get -3/2 on the outside

After you've done that you're just left with du
and du = xdx
so replace the x in the equation with du!

Answer 72

I'm sorry, but i have to go

-3/2 (integral sign) cos(u)du

integrating that gives you

(-3sin(u)) / (2) + C

= -3/2 * sin(1 - x^2) + C

Good luck!