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

Answer 1

Ezra Adams

That integral does not converge.

#1/(x^2-x-2)# is not continuous on the interval #[0,3]# (Discontinuous at #2#.)

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

#int_0^2 1/(x^2-x-2)dx + int_2^3 1/(x^2-x-2)dx#.

Neither integral converges.

To integrate #1/(x^2-x-2)# use partial fractions to get

#int 1/(x^2-x-2) dx = int (1/3 1/(x-2) - 1/3 1/(x+1))dx#

# = 1/3 lnabs(x-2)-1/3lnabs(x+1)#

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Answer 2

Ezra Adams

To find the integral of ( \int \frac{1}{x^2 - x - 2} , dx ) from 0 to 3, you can use the method of partial fraction decomposition followed by integration. After decomposing the rational function into partial fractions, integrate each term individually. Finally, evaluate the integral at the upper limit (3) and subtract the value of the integral at the lower limit (0).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.