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

anupjsebastian

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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
 
Lacking additional information, \(\displaystyle \int \dfrac{1}{x} dx = \ln(|x|) + C\)

Does this modify your musings?
 
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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?
 
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? ;)
 
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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