first, and looking only at the indefinite integral, we play with the denominator and see if it factors or squares
here we can say that #x^2 - x - 2 = (x-2)(x+1) qquad star# and it seems the lower integration limit is also a point at which the denominator is zero!!
so
#int 1/(x^2-x-2)dx#
#=int 1/( (x-2)(x+1))dx#
#=int A/(x-2) + B/(x+1)dx# , and we can use partial fractions
# A/(x-2) + B/(x+1) = (A(x+1) + B(x-2))/( (x-2)(x+1)) = (1)/( (x-2)(x+1))#
comparing numerators in the last two cases and....
...first set #x = 2 implies 3A = 1, A = 1/3#
...next set #x = -1 implies -3B = 1, B = -1/3#
#implies 1/3 int 1/(x-2) - 1/(x+1)dx#
#= 1/3 ( ln abs (x-2) - ln abs (x+1))#
#= 1/3 ln abs ((x-2)/(x+1))#
for the definite integral, this becomes
#[1/3 ln abs ((x-2)/(x+1))]_(-1)^1#
and if we just plug in the values...
#= 1/3[ ln abs ((1-2)/(1+1)) - ln abs ((-1-2)/(color(red)(-1+1)))]#
it might be clearer if the last term is expanded
# ln abs ((-1-2)/(color(red)(-1+1))) = ln abs ((-1-2) - ln abs (color(red)(-1+1))#
#ln 0 = -oo# and so the integral does not converge.
