Let #160; I=∫ x ^ 2 +x 6 ( x 2 ) ( x 1 ) #160; dx =∫( x+3 ) ( x 2 ) #160; ( x 2 ) ( x 1 ) #160; dx I=∫ x+3 x 1 dx =∫( x 1+4 x 1 #160 ...

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Integration by Partial Fractions

∫x2+x-6/x-2x-...

Question

$\int \frac{x^{2} + x - 6}{(x - 2) (x - 1)} d x =$

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Solution

Let $I = \int \frac{x^{2} + x - 6}{(x - 2) (x - 1)} d x = \int \frac{(x + 3) (x - 2)}{(x - 2) (x - 1)} d x$
$I = \int \frac{x + 3}{x - 1} d x = \int (\frac{x - 1 + 4}{x - 1}) d x = \int (1 + \frac{4}{x - 1}) d x$
So, $I = x + 4 l o g (x - 1) + C$

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