How do you find $\int \frac{x}{x^{2} - x - 2} d x$ $int x/(x^2-x-2)dx$ using partial fractions?

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How do you find $\int \frac{x}{x^{2} - x - 2} d x$ $int x/(x^2-x-2)dx$ using partial fractions?

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Answer 1 · 2016-10-02T22:41:33

$\int \frac{x}{x^{2} - x - 2} d x = \frac{2}{3} ln | x - 2 | + \frac{1}{3} ln | x + 1 | + C$ $int x/(x^2-x-2) dx = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C$

Explanation:

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

So:

$\int \frac{x}{x^{2} - x - 2} d x = \int \frac{2}{3 (x - 2)} + \frac{1}{3 (x + 1)} d x$ $int x/(x^2-x-2) dx = int 2/(3(x-2))+1/(3(x+1)) dx$

$\int \frac{x}{x^{2} - x - 2} d x = \frac{2}{3} ln | x - 2 | + \frac{1}{3} ln | x + 1 | + C$ $color(white)(int x/(x^2-x-2) dx) = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C$

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How do you find $\int \frac{x}{x^{2} - x - 2} d x$ $int x/(x^2-x-2)dx$ using partial fractions?

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