$∫ 2 x − 1/ ( x − 1 ) ( x + 2 ) ( x − 3 ) d x = A log | x − 1 |$ $+ B log | x + 2 | + C log | x − 3 | + K$ ,then A, B, and C are respectively

Question

$-\frac{1}{6}, -\frac{1}{3}, -\frac{1}{2}$

$\frac{1}{6}, \frac{1}{3}, -1$

$\frac{1}{6}, -\frac{1}{3}, \frac{1}{5}$

$-\frac{1}{6}, -\frac{1}{3}, \frac{1}{2}$

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