# SOLVEDIntegrate 1/(x-1) - 1/(x+1)

#### VinceW

$$\displaystyle \int \frac{1}{x - 1} - \frac{1}{x + 1} \operatorname{d} x$$

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.

#### Duke

You need modulus signs around (x-1) and (x+1).

#### Duke

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.

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