Question

\[\int\limits_{-1}^{0}{1 \over 2x-1}dx\]

Help with steps of integration and substitution.

Answer 1

Try a u substitution for 2x-1.

Answer 2

I first need to integrate it.

Answer 3

That's what you do with u substitution it allows to write everything in terms of u and du so it makes it easier to integrate it.

Answer 4

u=2x-1...du = 2dx.

Answer 5

Oh, yes... But I'm not sure how to do that with lns

Answer 6

Ok...so then you can rewrite the whole equation as 1/u (1/2du). Then pull the constant 1/2 out of the integral and integrate 1/u du.

Answer 7

Not sure how to do that either..?

Answer 8

Does the first part of substituting u for 2x-1 make sense?

Answer 9

Yes

Answer 10

And you know how we found what dx equals?

Answer 11

\[\int\limits \frac{1}{2x-1}dx\]
u=2x-1
\[\frac{du}{dx}= \frac{d}{dx}(2x-1)= 2\]
\[du=2dx\]

Answer 12

Yes, I understand... but how do you integrate 1/u du.

Answer 13

You just integrate everything in terms of u instead of x. So the integral of 1/u is ln(u).

Answer 14

And what does the "u" represent?

Answer 15

u=2x-1
we'll substitute 2x-1 later

Answer 16

Oh, Ok.

Answer 17

2x-1. So you can either resubstitute 2x-1 back in for u at this point and use your original limits of integration. Or you can plug your original limits into the u=2x-1 equation and use those new values in the ln(u) equation.

Answer 18

Can you show the steps of the substitution?

Answer 19

I'm not sure I know what the final equation is before I substitute?

Answer 20

Alright, you start with 1/2x-1. Then you let u =2x-1. Then you need to find dx by taking the derivative of 2x-1. So when you plug the u in for 2x-1 and 1/2 du in for dx...that is your new integrand.....1/u times 1/2du.

Answer 21

And then you just integrate 1/2(1/u)du...which gives you the antiderivative of ln(u).

Answer 22

How do you do that?

Answer 23

Do what exactly?

Answer 24

integrate 1/2(1/u)du

Answer 25

\[\int\limits_{-1}^{0} 1/(2x-1)\]
let u=2x-1
du=2dx
when x=-1, u=-3
when x=0, u=-1
now limit is from -3 to -1
= \[\int\limits_{-3}^{-1}(2/u) du\]
=2 ln(-3)-ln(-1)
=2 ln(-3/-1) because lna-lnb =ln(a/b)
=2 ln(3)
=\[\ln(3^{2})\]
= ln(9)
=2.1972

Answer 26

The 1/2 is a constant right?

Answer 27

yes

Answer 28

to write the answer here is really difficult

Answer 29

And you know that constants can pulled out of the integrand...right?

Answer 30

Yes

Answer 31

Great. So we pull it out...and all that's left to integrate is 1/u du. Do you know how to integrate 1/x dx?

Answer 32

It's Inx

But doesn't u stand for 2x-1

Answer 33

integrate (1/x dx)

Answer 34

Forget about that for a second...because we're not integrating for x anymore...we're integrating for u.

Answer 35

So, it would be lnU

Answer 36

Right! Now....what does U equal?

Answer 37

2x-1

Answer 38

So you can now put that back in for u in the lnu equation....but don't forget about the 1/2 we took out...which gets multiplied with the ln function.

Answer 39

Ok, so it's 1/2 * In(-1) - In(-3)

Answer 40

Yes...1/2ln(-1)-1/2ln(-3)

Answer 41

Ok, so the answer is -1/2ln3

Answer 42

Except you cant take logs of negative numbers!

Answer 43

Yes, that's the answer in the book :)

Answer 44

Thanks for helping me!

Answer 45

Ok....if they're cool with that. You bet. :)

Answer 46

Yes, they are :)