\(\int\displaystyle\frac{2x - 1}{(x - 1)(x + 2)(x - 3)}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle -\frac{1}{6}\log|x - 1| - \frac{1}{3}\log|x + 2| + \frac{1}{2}\log|x - 3| + C\)
Write the integrand in partial fractions:
\( \displaystyle \frac{2x - 1}{(x - 1)(x + 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{D}{x - 3} \)
\( 2x - 1 = A(x + 2)(x - 3) + B(x - 1)(x - 3) + D(x - 1)(x + 2) \)
Putting \(x = 1\): \(1 = A(3)(-2)\), so \(A = -\frac{1}{6}\).
Putting \(x = -2\): \(-5 = B(-3)(-5)\), so \(B = -\frac{1}{3}\).
Putting \(x = 3\): \(5 = D(2)(5)\), so \(D = \frac{1}{2}\).
Each term integrates to a logarithm:
\( \displaystyle \int \frac{2x - 1}{(x - 1)(x + 2)(x - 3)}\,dx = -\frac{1}{6}\log|x - 1| - \frac{1}{3}\log|x + 2| + \frac{1}{2}\log|x - 3| + C \)