Let' sstart with transforming the denominator.

x^{2} + x - 6 = (x+3)(x-2) (1)

Its reciprocal equals

1/[(x+3)(x-2] = (1/5)[(1/(x-2)) - (1/(x+3))] (2)

Thus, your function to be integrated is the differebce of two rational functions -

(1/5) x^{2}/(x-2) and (1/5)x^{2}/(x+3) (3)

Now let's transform these expressions to make them intregrable (without 1/5 coefficient).

In the first one subtract and add 4 to x^{2, }and in the second one subtract and add 9 to x^{2}. We have

(if we make use of the standard formula a^{2} - b^{2} = (a+b)(a-b) (notice: 4 = 2^{2} and 9 = 3^{2}))

x^{2}/(x-2) = (x^{2} - 4 + 4)/(x-2) = (x^{2} - 4)/(x-2) + 4 /(x-2) = x + 2 + [4/(x-2)] (4)

and

x^{2}/(x+3) = (x^{2} -9 +9)/(x+3) =(x^{2} -9)/(x+3) + 9/(x+3) = x - 3 + [9/(x+3)] (5)

If you subtract these two expressions you will obtain:

5 + 4/(x-2) - 9/(x+3)

Thus, your integral becomes:

∫ [x^{2}/(x^{2} + x - 6)] = (1/5) [∫5 dx + 4 ∫dx/(x-2) - 9 ∫ dx/(x+3)] =

(1/5) [ 5x + 4ln(x-2) - 9 ln(x+3)] + C .