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

In summary, the integration of 1/(x-1) can have two answers: y = ln(x-1) and y = ln(1-x), but the correct answer is y = ln|x-1| + C, taking into consideration the absolute value of the denominator.

## 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

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 :)

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

∫ 1/x dx = ln(|x|) +C, not ln(x) + C

## 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

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|

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

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.

## 1. Why does the integral of 1/(x-1) have two answers?

The integral of 1/(x-1) has two answers because the function 1/(x-1) has a singularity at x=1. This means that the function is undefined at x=1 and therefore, the integral has two possible values depending on which path is taken around the singularity.

## 2. Can you explain the concept of a singularity in terms of integrals?

A singularity is a point where a function is undefined or discontinuous. In the context of integrals, this means that the function is not continuous or differentiable at that point. This leads to the integral having two possible values because there are two different paths that can be taken around the singularity.

## 3. How do we know which path to take when evaluating the integral of 1/(x-1)?

The path taken when evaluating the integral of 1/(x-1) depends on the interval of integration. If the interval includes the singularity at x=1, then both possible values of the integral must be taken into account. If the interval does not include the singularity, then the integral has a single, unique value.

## 4. Can we avoid dealing with the singularity when evaluating the integral of 1/(x-1)?

No, we cannot avoid dealing with the singularity when evaluating the integral of 1/(x-1) because it is an essential part of the function. Trying to ignore the singularity would lead to incorrect results and violate the fundamental principles of integration.

## 5. How does the presence of a singularity affect the overall value of the integral of 1/(x-1)?

The presence of a singularity affects the overall value of the integral of 1/(x-1) by creating two possible values. This means that the integral is not a single, definite value and depends on the interval of integration. It is important to take the singularity into account when evaluating the integral to get an accurate result.

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