How do you find the integral of #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.

#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)#