Find $\displaystyle\int \dfrac{1}{(x-1)(x+1)}\,dx$.
A$\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$
B$\dfrac{1}{2}\ln\left|\dfrac{x+1}{x-1}\right| + C$
C$\ln|x^2 - 1| + C$
D$\dfrac{1}{x^2 - 1} + C$
Answer & Solution
Correct answer: A. $\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$
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.
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