Resolve frac{x + 1}{(x - 1)(x - 2)(x - 3)} in | vedclass

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(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,

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

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(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,we get:$x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$.For $x = 1$: $1 + 1 = A(1 - 2)(1 - 3) \implies 2 = A(-1)(-2) \implies 2 = 2A \implies A = 1$.For $x = 2$: $2 + 1 = B(2 - 1)(2 - 3) \implies 3 = B(1)(-1) \implies 3 = -B \implies B = -3$.For $x = 3$: $3 + 1 = C(3 - 1)(3 - 2) \implies 4 = C(2)(1) \implies 4 = 2C \implies C = 2$.Thus,the partial fraction is $\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3}$.

Resolve $\frac{x + 1}{(x - 1)(x - 2)(x - 3)}$ into partial fractions.

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