Given #160; I=∫ x (x 1)(x 2) #160; dx x (x 1)(x 2) = A x 1 + B x 2 #160; #160; #160; #160; #160; #160; #160; #160; ...

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Solving Equations with Variables on Both Sides

Find ∫xx-1x...

Question

Find $\int \frac{x}{(x - 1) (x - 2)} d x$ .

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Solution

Given $I = \int \frac{x}{(x - 1) (x - 2)} d x$
$\frac{x}{(x - 1) (x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$
$= \frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)}$
$A + B = 1, - 2 A - B = 0$
$A - 2 A = 1, 2 A = - B$
$A = - 1$
$B = 2$
$⟹ I = \int [\frac{- 1}{x - 1} + \frac{2}{x - 2}] d x$
$= \int \frac{2 d x}{x - 2} - \int \frac{d x}{x - 1}$
$= 2 ln | x - 2 | - ln | x - 1 |$

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