Verified answer

First factor the denominator:

x² - x + 2 = (x - 2)(x + 1)

Now use the method of partial fractions:

make 1 / (x - 2)(x + 1) into the form A/(x - 2) + B/(x + 1)

1/ (x-2)(x+1) = A/(x - 2) + B/(x + 1)

Multiply both sides by (x - 2)(x + 1)

1 = A(x + 1) + B(x - 2)

Pick strategic values of x so that it will cancel one term on the right and let find one of A and B, then choose another value of x that finds the other of A and B.

Let x = -1.

1 = A(-1 + 1) + B(-1 - 2)

1 = 0 - 3B

B = -1/3

Let x = 2:

1 = A(2 + 1) + B(2 - 2)

1 = 3A + 0

A = 1/3

So Now we have:

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

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

So the original integrand:

1/(x² - x + 2)

becomes

1 / 3(x - 2) - 1 / 3(x + 1)

So integrate:

int [ 1 / 3(x - 2) - 1 / 3(x + 1) ] dx

= 1/3 log(x - 2) - 1/3 log(x + 1) + C