Find $\displaystyle\int \dfrac{1}{(x-1)(x+1)}\,dx$.

Use partial fractions. Write $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$. Multiplying out: $1 = A(x+1) + B(x-1)$ Plug $x = 1$: $1 = 2A \Rightarrow A = \dfrac{1}{2}$. Plug $x = -1$: $1 = -2B \Rightarrow B = -\dfrac{1}{2}$. So the integral becomes: $\displaystyle\int \dfrac{1/2}{x-1}\,dx - \int \dfrac{1/2}{x+1}\,dx = \dfrac{1}{2}\left(\ln|x-1| - \ln|x+1|\right) + C = \dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$. **Watch the sign.** Option B has the ratio inverted. Differentiating option A confirms it gives the original integrand.