integrate (2-x)/(x-1)

[TanisHalfElven]

can anyone help here.
i can put in my ti-89 and get the answer but i can't figure out how to do it.
i tried substitution but i can't get it right.
any help appreciated.

 

[fralexandr]

foil it nvm didn't see the divide sign

 

[TanisHalfElven]

Originally posted by: fralexandr
foil it

what ?

 

[thesurge]

Hint:
x/(x-1)=1/(x-1)+1

 

[RagingBITCH]

You fail at math.

 

[invidia]

Originally posted by: thesurge
Hint:
x/(x-1)=[1/(x-1)]+1

 

[George P Burdell]

Text

 

[Cheesetogo]

Hint

 

[TanisHalfElven]

Originally posted by: RagingBITCH
You fail at math.

yeah i know. but i'm in a diffeq class. i haven't practised stupid confusing integrals in a long time.

 

[nycxandy]

Originally posted by: invidia

Originally posted by: thesurge
Hint:
x/(x-1)=[1/(x-1)]+1

Redundant.

 

[gorcorps]

http://integrals.wolfram.com/index.jsp

everybody's best friend

BTW it may help to break it up into parts: [2/(x-1)]-[x/(x-1)] then go from there

 

[TanisHalfElven]

Originally posted by: Cheesetogo
Hint

thanks. that pretty cool.

 

[Eeezee]

Simple u substitution should do the job. It's one of the most useful and easy to use integration tools. You can make up a variable (u) that equals whatever you want! I usually use it to eliminate the denominator. Once you get good at using it, you can do these integrals in your head (I can hear the undergrads screaming 'impossible!' but it's not hard if you're willing to try).

u = x-1, dx = du, x = u+1

Then you get (2 - (u + 1))/u = (1-u) / u = (1/u) - 1

That's a ln(u) - u + C

ie the answer ln(x-1) - x + 1 + C

If you're not completely sure, it's always good to check your answer. Take the derivative of what we've found and you get
(1/(x-1)) - 1
= (1/(x-1)) - ((x-1)/(x-1))
= (1 - x + 1) / (x-1)
= (2-x) / (x-1)
Sweet, we got the argument back, so the answer is

ln(x-1) - x + 1 + C

I'm getting my PhD in physics, I want to become a teacher

 

[Safeway]

Originally posted by: Eeezee
Simple u substitution should do the job. It's one of the most useful integration tools.

u = x-1, dx = du, x = u+1

Then you get (2 - (u + 1))/u = (1-u) / u = (1/u) - 1

That's a ln(u) - u + C

ie the answer ln(x-1) - x + 1 + C

If you're not completely sure, it's always good to check your answer. Take the derivative of what we've found and you get
(1/(x-1)) - 1
= (1/(x-1)) - ((x-1)/(x-1))
= (1 - x + 1) / (x-1)
= (2-x) / (x-1)
Sweet, we got the argument back, so the answer is

ln(x-1) - x + 1 + C

I'm getting my PhD in physics, I want to become a teacher

But I thought that, "Those who can't, teach." You are not following that logic.

 

[Eeezee]

Originally posted by: Safeway

Originally posted by: Eeezee
Simple u substitution should do the job. It's one of the most useful integration tools.

u = x-1, dx = du, x = u+1

Then you get (2 - (u + 1))/u = (1-u) / u = (1/u) - 1

That's a ln(u) - u + C

ie the answer ln(x-1) - x + 1 + C

If you're not completely sure, it's always good to check your answer. Take the derivative of what we've found and you get
(1/(x-1)) - 1
= (1/(x-1)) - ((x-1)/(x-1))
= (1 - x + 1) / (x-1)
= (2-x) / (x-1)
Sweet, we got the argument back, so the answer is

ln(x-1) - x + 1 + C

I'm getting my PhD in physics, I want to become a teacher

But I thought that, "Those who can't, teach." You are not following that logic.

I'm sure physics, chemistry and biology researchers at universities around the world would also have something to say about that. Many of the brightest minds were teachers. Feynman and Einstein were both teachers, and they clearly "can."

"Those who can't, teach" really only applies to business work environments. I had a math teacher in middle school who hated math and failed out of the business world - obviously he fits the phrase. I had a math teacher in high school who had a PhD in computer science and she owned several patents before she got her BS - obviously she does not fit the old phrase.

I'd teach high school if there were more money in it, but the good pay is in higher learning (community college or university)

 

[TanisHalfElven]

Originally posted by: Eeezee
Simple u substitution should do the job. It's one of the most useful and easy to use integration tools. You can make up a variable (u) that equals whatever you want! I usually use it to eliminate the denominator. Once you get good at using it, you can do these integrals in your head (I can hear the undergrads screaming 'impossible!' but it's not hard if you're willing to try).

u = x-1, dx = du, x = u+1

Then you get (2 - (u + 1))/u = (1-u) / u = (1/u) - 1

That's a ln(u) - u + C

ie the answer ln(x-1) - x + 1 + C

If you're not completely sure, it's always good to check your answer. Take the derivative of what we've found and you get
(1/(x-1)) - 1
= (1/(x-1)) - ((x-1)/(x-1))
= (1 - x + 1) / (x-1)
= (2-x) / (x-1)
Sweet, we got the argument back, so the answer is

ln(x-1) - x + 1 + C

I'm getting my PhD in physics, I want to become a teacher

thanks.
good luck in your PhD. if you don't already read phdcomics

 

[Syringer]

Originally posted by: nycxandy

Originally posted by: invidia

Originally posted by: thesurge
Hint:
x/(x-1)=[1/(x-1)]+1

Redundant.

Not really, here's a method without using integration:

(2-x)/(x-1)

= 2/(x-1) - x/(x-1)

= 2/(x-1) - (x+(-1+1))/(x-1) --> Added one and subtracted one to the numerator

= 2/(x-1) - (1+(x-1))/(x-1)

= 2/(x-1) - 1/(x-1) - (x-1)/(x-1)

= (2-1)/(x-1) - 1

= 1/(x-1) - 1

Take the integral of that...

= ln(x-1) - x + C

Makes more sense if you write it out rather than read it off the computer screen..

 

[Eeezee]

Originally posted by: tanishalfelven

Originally posted by: Eeezee
Simple u substitution should do the job. It's one of the most useful and easy to use integration tools. You can make up a variable (u) that equals whatever you want! I usually use it to eliminate the denominator. Once you get good at using it, you can do these integrals in your head (I can hear the undergrads screaming 'impossible!' but it's not hard if you're willing to try).

u = x-1, dx = du, x = u+1

Then you get (2 - (u + 1))/u = (1-u) / u = (1/u) - 1

That's a ln(u) - u + C

ie the answer ln(x-1) - x + 1 + C

If you're not completely sure, it's always good to check your answer. Take the derivative of what we've found and you get
(1/(x-1)) - 1
= (1/(x-1)) - ((x-1)/(x-1))
= (1 - x + 1) / (x-1)
= (2-x) / (x-1)
Sweet, we got the argument back, so the answer is

ln(x-1) - x + 1 + C

I'm getting my PhD in physics, I want to become a teacher

thanks.
good luck in your PhD. if you don't already read phdcomics

Thanks, I read that and xkcd when I'm trying to procrastinate