How do you use partial fractions to find the integral $\int \frac{2 x^{2} - 2 x + 3}{x^{3} - x^{2} - x - 2} d x$ $int (2x^2-2x+3)/(x^3-x^2-x-2)dx$ ?

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How do you use partial fractions to find the integral $\int \frac{2 x^{2} - 2 x + 3}{x^{3} - x^{2} - x - 2} d x$ $int (2x^2-2x+3)/(x^3-x^2-x-2)dx$ ?

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Answer 1 · 2016-12-23T00:55:17

By substituting small numbers as a guess, $x = 2$ $x=2$ makes the denominator zero, so the partial fraction expression, by the cover up rule, is
$\frac{\frac{7}{7}}{x - 2} + \frac{A x + B}{x^{2} + x + 1}$ $(7/7)/(x-2)+(Ax+B)/(x^2+x+1)$

Explanation:

Then equating powers of $x$ $x$ you get $A = 1$ $A=1$ , $B = - 1$ $B=-1$ . The second term is now linear divided by quadratic which is routine.

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How do you use partial fractions to find the integral $\int \frac{2 x^{2} - 2 x + 3}{x^{3} - x^{2} - x - 2} d x$ $int (2x^2-2x+3)/(x^3-x^2-x-2)dx$ ?

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How do you use partial fractions to find the integral $\int \frac{2 x^{2} - 2 x + 3}{x^{3} - x^{2} - x - 2} d x$ $int (2x^2-2x+3)/(x^3-x^2-x-2)dx$ ?