$\frac{(x + 1)}{(x - 1) (x - 2) (x - 3)} =$

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

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

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

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None of these

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Solution

The correct option is A
$\frac{1}{(x - 1)} - \frac{3}{(x - 2)} + \frac{2}{(x - 3)}$

Explanation for the correct option:
Step 1: Rewrite the given expression as a sum of expression
Let's consider,
$\frac{(x + 1)}{(x - 1) (x - 2) (x - 3)} = \frac{A}{(x - 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 3)}$
$\Rightarrow$ $\frac{(x + 1)}{(x - 1) (x - 2) (x - 3)} = \frac{A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)}{(x - 1) (x - 2) (x - 3)}$
Comparing the numerator, we get
$x + 1 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)$
$\Rightarrow$ $x + 1 = A (x^{2} - 5 x + 6) + B (x^{2} - 4 x + 3) + C (x^{2} - 3 x + 2)$
$\Rightarrow$ $x + 1 = (A + B + C) x^{2} + (- 5 A - 4 B - 3 C) x + (6 A + 3 B x + 2 C)$
Step 2: Comparing the coefficients
$A + B + C = 0$ $. . . . . (1)$
$- 5 A - 4 B - 3 C = 1$ $. . . . . (2)$
$6 A + 3 B + 2 C = 1$ $. . . . . (3)$
From equation $(1)$ , we get
$A = - B - C$
Substitute this value in equations $(2)$ and $(3)$ , we get
$- 5 (- B - C) - 4 B - 3 C = 1$ and $6 (- B - C) + 3 B + 2 C = 1$
$\Rightarrow$ $B + 2 C = 1$ and $- 3 B - 4 C = 1$

Substitute $B = 1 - 2 C$ in $- 3 B - 4 C = 1$ , we get
$- 3 (1 - 2 C) - 4 C = 1$
$\Rightarrow$ $- 3 + 6 C - 4 C = 1$
$\Rightarrow$ $2 C = 4$
$\Rightarrow$ $C = 2$
Solving the equations by substituting $C = 2$ , we get
$A = 1, B = - 3$
Thus,
$\frac{(x + 1)}{(x - 1) (x - 2) (x - 3)} = \frac{1}{(x - 1)} + \frac{- 3}{(x - 2)} + \frac{2}{(x - 3)}$

Hence, the correct option is $A) \frac{1}{(x - 1)} - \frac{3}{(x - 2)} + \frac{2}{(x - 3)}$

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