How do you find the integral of $\int \frac{1}{x^{2} - x - 2} d x$ $int (1)/(x^2 - x - 2) dx$ from 0 to 3?

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How do you find the integral of $\int \frac{1}{x^{2} - x - 2} d x$ $int (1)/(x^2 - x - 2) dx$ from 0 to 3?

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Answer 1 · 2015-10-10T20:31:06

That integral does not converge.

Explanation:

$\frac{1}{x^{2} - x - 2}$ $1/(x^2-x-2)$ is not continuous on the interval $[0, 3]$ $[0,3]$ (Discontinuous at $2$ $2$ .)

When we have learned about improper integrals we can attempt to evaluate

$\int_{0}^{2} \frac{1}{x^{2} - x - 2} d x + \int_{2}^{3} \frac{1}{x^{2} - x - 2} d x$ $int_0^2 1/(x^2-x-2)dx + int_2^3 1/(x^2-x-2)dx$ .

Neither integral converges.

To integrate $\frac{1}{x^{2} - x - 2}$ $1/(x^2-x-2)$ use partial fractions to get

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

$= \frac{1}{3} ln | x - 2 | - \frac{1}{3} ln | x + 1 |$ $= 1/3 lnabs(x-2)-1/3lnabs(x+1)$

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How do you find the integral of $\int \frac{1}{x^{2} - x - 2} d x$ $int (1)/(x^2 - x - 2) dx$ from 0 to 3?

1 Answer

Explanation:

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How do you find the integral of int (1)/(x^2 - x - 2) dx∫1x2−x−2dxint (1)/(x^2 - x - 2) dx from 0 to 3?

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How do you find the integral of $\int \frac{1}{x^{2} - x - 2} d x$ $int (1)/(x^2 - x - 2) dx$ from 0 to 3?