Question

du=-3 dx then we get dx =______?

Answer 1

i though it would be -1/3

Answer 2

but its not

Answer 3

It is.

Answer 4

its not it....i dont get it...

Answer 5

Or rather it is:

\[du = -3 dx \implies dx = -\frac{1}{3}du\]

Answer 6

Yes, I'm afraid it is.

Answer 7

ok thanks polpak : )

Answer 8

polpak i need to complet this step: \[u= 2-3x \int\limits dx/2-3x= \int\limits 1/u (-1/3 du) = (1/3) \int\limits 1/u \]

Answer 9

du

Answer 10

I don't understand what you have there...

Answer 11

Are you trying to solve

\[\int \frac{dx}{2-3x}\] ?

Answer 12

that becomes 1/u...but i need to put it in the final answer as (-1/3) (1/u) + C

Answer 13

The integral of 1/u du is the natural log of u plus a constant.

Answer 14

so the natural log of 2 -3x is.....

Answer 15

That's it.

Answer 16

Unless you have some limits you can plug in for your x.

Answer 17

so it would become (-1/3)(2-3x)+C

Answer 18

no.

Answer 19

\[-\frac{1}{3}ln(2-3x) + C\]

Answer 20

there's no limits

Answer 21

o so then what i do with it? do i take the ln of -1/3?

Answer 22

No. you don't do anything. that's all there is.

Answer 23

\[\text{Let } u = 2-3x\]\[\implies du = -3dx\]\[\implies dx = -\frac{1}{3}du\]\[\implies \int \frac{dx}{2-3x} = \int \frac{1}{u}(-\frac{1}{3} du)\]\[ =-\frac{1}{3}ln(2-3x) + C\]

Answer 24

this is what its asking me to do : -1/3 ln\[\left| u \right|\] + C =_______

Answer 25

Yes fine, technically the () should be ||

Answer 26

and i put 2-3x but its wrong

Answer 27

yes, because -1/3ln|u| + C is not 2-3x.

Answer 28

Just replace the u with what u equals.

Answer 29

but you have to write the whole thing.

Answer 30

2+3x?

Answer 31

No

Answer 32

i am doing this online and the system telling me it wrong when i put (-1/3)ln(2-3x)+C....

Answer 33

well the answer is {-(ln(2-3x))/3}+C

Answer 34

Those two things are the same.

Answer 35

So your system is dumb if it can't tell that they are equivalent.

Answer 36

i know right : (