How do you evaluate the integral $\int \frac{2 x^{2} + x - 5}{(x - 3) (x + 2)}$ $int (2x^2+x-5)/((x-3)(x+2))$ ?

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How do you evaluate the integral $\int \frac{2 x^{2} + x - 5}{(x - 3) (x + 2)}$ $int (2x^2+x-5)/((x-3)(x+2))$ ?

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Answer 1 · 2018-02-19T12:21:14

$\int \frac{2 x^{2} + x - 5}{x^{2} - x - 6} \cdot d x = 2 x + 2 ln (x - 3) - ln (x + 2) + C$ $int (2x^2+x-5)/(x^2-x-6)*dx=2x+2ln(x-3)-ln(x+2)+C$

Explanation:

$\int \frac{2 x^{2} + x - 5}{x^{2} - x - 6} \cdot d x$ $int (2x^2+x-5)/(x^2-x-6)*dx$

= $\int 2 d x$ $int 2dx$ + $\int \frac{x + 7}{x^{2} - x - 6} \cdot d x$ $int (x+7)/(x^2-x-6)*dx$

= $2 x + \int \frac{x + 7}{(x - 3) \cdot (x + 2)} \cdot d x$ $2x+int (x+7)/((x-3)*(x+2))*dx$

= $2 x + \int \frac{2 x + 4}{(x - 3) (x + 2)} \cdot d x$ $2x+int (2x+4)/((x-3)(x+2))*dx$ - $\int \frac{x - 3}{(x - 3) (x + 2)} \cdot d x$ $int (x-3)/((x-3)(x+2))*dx$

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

= $2 x + 2 ln (x - 3) - ln (x + 2) + C$ $2x+2ln(x-3)-ln(x+2)+C$

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How do you evaluate the integral $\int \frac{2 x^{2} + x - 5}{(x - 3) (x + 2)}$ $int (2x^2+x-5)/((x-3)(x+2))$ ?

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Explanation:

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How do you evaluate the integral ∫2x2+x−5(x−3)(x+2)int (2x^2+x-5)/((x-3)(x+2))?

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How do you evaluate the integral $\int \frac{2 x^{2} + x - 5}{(x - 3) (x + 2)}$ $int (2x^2+x-5)/((x-3)(x+2))$ ?