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

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

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

x2 + 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) x2/(x-2)     and   (1/5)x2/(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 x2, and in the second one subtract and add 9 to x2. We have

(if we make use of  the standard formula  a2 - b2 = (a+b)(a-b)  (notice: 4 = 22  and 9 = 32))

x2/(x-2) = (x2 - 4 + 4)/(x-2) = (x2 - 4)/(x-2) + 4 /(x-2) = x + 2 + [4/(x-2)]      (4)

and

x2/(x+3) = (x2 -9 +9)/(x+3) =(x2 -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)