How do you integrate $\int \frac{x}{x^{3} - 2 x^{2} + x}$ $int x/ (x^3-2x^2+x)$ using partial fractions?

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How do you integrate $\int \frac{x}{x^{3} - 2 x^{2} + x}$ $int x/ (x^3-2x^2+x)$ using partial fractions?

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Answer 1 · 2018-06-10T21:14:33

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

Explanation:

Simplify the integrand:

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

Factorize the denominator:

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

Then:

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

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How do you integrate $\int \frac{x}{x^{3} - 2 x^{2} + x}$ $int x/ (x^3-2x^2+x)$ using partial fractions?

1 Answer

Explanation:

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How do you integrate $\int \frac{x}{x^{3} - 2 x^{2} + x}$ $int x/ (x^3-2x^2+x)$ using partial fractions?