I am trying to understand what's best to start this problem.

## Should I use U-substitution for finding the Integral of X^2/(x^2 + x-6)

# 4 Answers

You can use partial fraction.

x^2/(x^2 + x-6)

= (x^2+x-6-x+6)/[(x+3)(x-2)]

= 1 - (x-6)/[(x+3)(x-2)]

Let (x-6)/[(x+3)(x-2)] = A/(x+3) + B/(x-2)

Multiply both sides by x^2+x-6,

- (x-6) = A(x-2) + B(x+3)

Let x = -3,

9 = -5A => A = -9/5

Let x = 2,

4 = 5B => B = 4/5

So, the integral becomes

∫1 - (9/5)/(x+3) + (4/5)/(x-2) dx

= x - (9/5)ln|x+3| + (4/5)ln|x-2| + c <==Answer

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 .

Please try integrating using partial fractions; factorize the denominator first.

This is a partial fractions problem. But notice that it's an improper fraction (because the highest power in the numerator is as large as the highest power in the denominator), so before you can apply the partial fractions technique, you need to use long
polynomial division to break it down to 1 - (x-6)/(x^{2}+x-6). Now it's the sum of two terms. The first term, 1, integrates immediately to x. The second term (which includes the minus sign that precedes it) is now a proper fraction (the highest
power in the numerator is lower than the highest power in the denominator), so now you can go ahead and use partial fractions (as demonstrated in Robert J's answer) to break it down. Once you've broken that fraction down to its two component fractions, you
can integrate them separately. Each one integrates as a natural logarithm (ln); if you're not confident about your integration of each of these, you should use u-substitution on each to clarify the situation.