Using the improper integrals

a) integral going 0 to 3 1 / (x^2+x-2)dx

b) integral going 0 to 3 (x-1) / (x^2+x-2)dx

a)
∫[0,3] 1 / (x² + x - 2) dx

Notices that if we set the denominator equal to zero we get:
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 and x = 1

SInce x = 1 lies on the interval [0,3] that we are integrating over, this makes it an improper integral since there would be a vertical asymptote on the graph. Thus we must break up the integral:
∫[0,3] 1 / [(x + 2)(x - 1)] dx
= ∫[0,1] 1 / [(x + 2)(x - 1)] dx + ∫[1,3] 1 / [(x + 2)(x - 1)] dx

Use decomposition by partial fractions to get:
∫[0,1] [1 / (3(x-1))] - [1/ (3(x+2))] dx + ∫[1,3] [1/ (3(x-1))] - [1/ (3(x+2))]
= [(1/3)ln|x - 1| - (1/3)ln|x+2|]from x=0 to 1 + [(1/3)ln|x-1| - (1/3)ln|x+2|] from x=1 to 3

When we plug in x = 1, we will actually take the limit as x approaches 1 becase of the horizontal asymptote mentioned earlier:
= (1/3)lim(x-->1) [ln|x-1| - ln|x+2|] - (1/3)(ln(1) - ln(3)) + (1/3)(ln(2) - ln(5)) - (1/3)lim(x-->1) [ln|x - 1| - ln|x+2|]
= ∞

Thus the integral does not converge.



b)
∫[0,3] (x - 1) / (x² + x - 2) dx
= ∫[0,3] (x - 1) / [(x + 2)(x - 1)] dx
= ∫[0,3] 1 / (x + 2) dx
= ln|x + 2| from x=0 to 3
= ln(5) - ln(2)
= ln(5/2)