Integral of 1/(x-1) has two answers?

[RadiantL]

Homework Statement

Hi so I'm integrating 1/(x-1) but I think it may have two answers and I'm not sure if I'm right or wrong

Homework Equations

y = 1/(x-1)

The Attempt at a Solution

well if you integrate this function you would get

y = ln(x-1) (using u substituition)
I think it makes sense, you take the derivative and you get the same thing back.

now... I think there maybe another answer? it looks like this

y = ln(1-x)

if you take the derivative of this function you would get

y = 1/(1-x) χ -1
which would simplify to
y = 1/(x-1)

so yeah... I'm not sure which is the answer y = ln(x-1) or y = ln(1-x)

I believe I am missing something, thanks for the help :)

 

[tiny-tim]

Hi RadiantL!

ln(x) isn't defined if x < 0

∫ 1/x dx = ln(|x|) +C, not ln(x) + C

 

[Dalek1099]

Homework Statement

Hi so I'm integrating 1/(x-1) but I think it may have two answers and I'm not sure if I'm right or wrong

Homework Equations

y = 1/(x-1)

The Attempt at a Solution

well if you integrate this function you would get

y = ln(x-1) (using u substituition)
I think it makes sense, you take the derivative and you get the same thing back.

now... I think there maybe another answer? it looks like this

y = ln(1-x)

if you take the derivative of this function you would get

y = 1/(1-x) χ -1
which would simplify to
y = 1/(x-1)

so yeah... I'm not sure which is the answer y = ln(x-1) or y = ln(1-x)

I believe I am missing something, thanks for the help :)

its ln|x-1| + C and as the previous replier said don't forget to put the absolute value symbol around x-1 because the absolute value is here f(y)=f(-y) which means that ln|x-1|=ln|1-x|

 

[RadiantL]

Oh I see, that makes sense haha. Thanks tiny-tim and Dalek1099!

 

[Ray Vickson]

Homework Statement

Hi so I'm integrating 1/(x-1) but I think it may have two answers and I'm not sure if I'm right or wrong

Homework Equations

y = 1/(x-1)

The Attempt at a Solution

well if you integrate this function you would get

y = ln(x-1) (using u substituition)
I think it makes sense, you take the derivative and you get the same thing back.

now... I think there maybe another answer? it looks like this

y = ln(1-x)

if you take the derivative of this function you would get

y = 1/(1-x) χ -1
which would simplify to
y = 1/(x-1)

so yeah... I'm not sure which is the answer y = ln(x-1) or y = ln(1-x)

I believe I am missing something, thanks for the help :)

\int \frac{1}{1-x} \, dx = - \ln(1-x), because when you defferentiate the right-hand-side you get back the integrand.