alleryUsing partial fractions, $\displaystyle\int \dfrac{1}{(x-1)(x-2)}\,dx$ equals:
A$\log|(x-1)(x-2)| + C$
B$\dfrac{1}{2}\log\left|\dfrac{x-2}{x-1}\right| + C$
C$\log\left|\dfrac{x-2}{x-1}\right| + C$
D$\log\left|\dfrac{x-1}{x-2}\right| + C$
Answer & Solution
Correct answer: C. $\log\left|\dfrac{x-2}{x-1}\right| + C$
$\dfrac{1}{(x-1)(x-2)} = \dfrac{-1}{x-1} + \dfrac{1}{x-2}$ (since $A = 1/(1-2) = -1$ and $B = 1/(2-1) = 1$).
Integrating term by term gives $-\log|x-1| + \log|x-2| + C = \log\left|\dfrac{x-2}{x-1}\right| + C$.
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
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