$Evaluate \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

Easy

A

$= - \frac{1}{2} \log | x - 1 | - \frac{1}{6} \log | x + 2 | + \frac{1}{3} \log | x - 3 | + C$
B

$= - \frac{1}{3} \log | x - 1 | - \frac{1}{6} \log | x + 2 | + \frac{1}{2} \log | x - 3 | + C$
C

$= - \frac{1}{6} \log | x - 1 | - \frac{1}{3} \log | x + 2 | + \frac{1}{2} \log | x - 3 | + C$
D

None of these

Solution

$S i n c e a l l t h e f a c t o r s i n t h e d e n o m i n a t o r a r e l i n e a r, w e h a v e$
$\begin{array}{l} \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x \\ f o r m u l a : I f \frac{f (x)}{g (x)} = \frac{A_{1}}{x - a_{1}} + \frac{A_{2}}{x - a_{2}} + \frac{A_{3}}{x - a_{3}} + . . . . . . . \frac{A_{n}}{x - a_{n}} w h e r e g (x) = (x - a_{1}) (x - a_{2}) (x - a_{3}) . . . . (x - a_{n}) \\ t h e n A_{1} = \frac{f (a_{1})}{(a_{1} - a_{2}) (a_{1} - a_{3}) . . . (a_{1} - a_{n})}, A_{1} = \frac{f (a_{2})}{(a_{2} - a_{1}) (a_{2} - a_{3}) . . . (a_{2} - a_{n})}, . . . . . . . . A_{n} = \frac{f (a_{n})}{(a_{n} - a_{1}) (a_{n} - a_{2}) . . . (a_{n} - a_{n - 1})} \\ = \int [\frac{1}{(x - 1) (3) (- 2)} + \frac{- 5}{(- 3) (x + 2) (- 5)} + \frac{5}{(2) (5) (x - 3)}] d x \\ = - \frac{1}{6} \log | x - 1 | - \frac{1}{3} \log | x + 2 | + \frac{1}{2} \log | x - 3 | + C \end{array}$

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