Let' sstart with transforming the denominator.
x^{2} + x  6 = (x+3)(x2) (1)
Its reciprocal equals
1/[(x+3)(x2] = (1/5)[(1/(x2))  (1/(x+3))] (2)
Thus, your function to be integrated is the differebce of two rational functions 
(1/5) x^{2}/(x2) 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)(ab) (notice: 4 = 2^{2} and 9 = 3^{2}))
x^{2}/(x2) = (x^{2}  4 + 4)/(x2) = (x^{2}  4)/(x2) + 4 /(x2) = x + 2 + [4/(x2)] (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/(x2)  9/(x+3)
Thus, your integral becomes:
∫ [x^{2}/(x^{2} + x  6)] = (1/5) [∫5 dx + 4 ∫dx/(x2)  9 ∫ dx/(x+3)] =
(1/5) [ 5x + 4ln(x2)  9 ln(x+3)] + C .
7/24/2013

Grigori S.