SOLVED Integrate 1/(x-1) - 1/(x+1)

Jun 2010
59
4
\(\displaystyle \int \frac{1}{x - 1} - \frac{1}{x + 1} \operatorname{d} x\)

At first glance, the intuitive answer is simply \(\displaystyle \ln (x-1) - \ln (x+1) = \ln\frac{x-1}{x+1}\), however computer algebra systems give me a slightly different answer: \(\displaystyle \ln (1 - x) - \ln (x+1)\).

Obviously, the first term was negated. I assume this has something to do with the fact that logarithms only have real outputs with non-negative inputs.

Could someone please explain this?
 
May 2011
169
6
You need modulus signs around (x-1) and (x+1).
 
May 2011
169
6
If (x-1) is negative the abs(x-1) = 1-x. So you are both wrong because you have each assumed what the sign of it is.
 
Dec 2009
3,120
1,342
From the perspective of taking derivatives...

\(\displaystyle \frac{d}{dx}ln(1-x)=\frac{1}{1-x}(-1)=\frac{1}{x-1}\)
 
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