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Thread: Integration of 1/(-x+1) and -1/(x-1): Why the different results?

  1. #1

    Question Integration of 1/(-x+1) and -1/(x-1): Why the different results?

    Both these integrals should yield the same result as -x+1 = -(x-1)

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

    however one integrates to -ln(1-x) and the other to -ln(x-1)

    x-1 is not equal to 1-x for all values except x = 1

    so is the integral only defined for x = 1?

    but it is possible to plot the integral over multiple values...

    could someone please explain what is actually happening here..

    thanks

  2. #2
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    Lacking additional information, [tex]\int \dfrac{1}{x} dx = \ln(|x|) + C[/tex]

    Does this modify your musings?
    Last edited by Subhotosh Khan; 02-20-2018 at 12:16 PM. Reason: typo
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Thanks for the reply, but that wasn't my question.

    It confuses me why moving the minus from the denominator to the numerator gives a different integral as the answer

    -1/(x-1) integrates to -ln(x-1)

    1/-(x-1) integrates to -ln(x-1)

    x-1 is not equal to 1-x.

    Why does this occur?

  4. #4
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by anupjsebastian View Post
    Thanks for the reply, but that wasn't my question.

    It confuses me why moving the minus from the denominator to the numerator gives a different integral as the answer

    -1/(x-1) integrates to -ln(x-1)

    1/-(x-1) integrates to -ln(x-1)

    x-1 is not equal to 1-x.

    Why does this occur?
    How would the absolute value (pointed out in the first reply) affect these expressions?

  5. #5

    Yeah makes sense with the absolute value... sorry my bad I didn't get the first reply.

    thank you tkhunny and stapel .

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