How do you integrate $\int \frac{3 x - 4}{(x - 1) (x + 3) (x - 6)}$ $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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How do you integrate $\int \frac{3 x - 4}{(x - 1) (x + 3) (x - 6)}$ $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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Answer 1 · 2016-04-01T03:58:05

Partial fraction decomposition is:
$\frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{x - 6}$ $A/(x-1)+B/(x+3)+C/(x-6)$
Cover up rule or simultaneously solving by equating coefficients yields $A = \frac{1}{20}$ $A=1/20$ , $B = - \frac{13}{36}$ $B=-13/36$ and $C = \frac{14}{45}$ $C=14/45$

$\int \frac{3 x - 4}{(x - 1) (x + 3) (x - 6)} d x$ $int(3x-4)/((x-1)(x+3)(x-6))dx$
$= \frac{1}{20} \int \frac{1}{x - 1} d x - \frac{13}{36} \int \frac{1}{x + 3} d x + \frac{14}{45} \int (\frac{1}{x} - 6) d x$ $=1/20int1/(x-1)dx-13/36int1/(x+3)dx+14/45int(1/x-6)dx$
$= \frac{1}{20} ln | x - 1 | - \frac{13}{36} ln | x + 3 | + \frac{14}{45} ln | x - 6 | + c$ $=1/20ln|x-1|-13/36ln|x+3|+14/45ln|x-6|+c$

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How do you integrate $\int \frac{3 x - 4}{(x - 1) (x + 3) (x - 6)}$ $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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How do you integrate $\int \frac{3 x - 4}{(x - 1) (x + 3) (x - 6)}$ $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?