Integral x^2/(x-3)(x+2)^2?

Hello,

partial fraction decomposition:

x² /[(x - 3)(x + 2)²] = A/(x - 3) + B/(x + 2) + C/(x + 2)²

(the term B/(x + 2) is needed too since (x + 2) is squared)

((x - 3)(x + 2)²: common denominator)

x² /[(x - 3)(x + 2)²] = [A(x + 2)² + B(x + 2)(x - 3) + C(x - 3)] /[(x - 3)(x + 2)²]

(equating numerators)

x² = A(x + 2)² + B(x + 2)(x - 3) + C(x - 3)

x² = A(x² + 4x + 4) + B(x² - 3x + 2x - 6) + Cx - 3C

x² = Ax² + 4Ax + 4A + B(x² - x - 6) + Cx - 3C

x² = Ax² + 4Ax + 4A + Bx² - Bx - 6B + Cx - 3C

x² = (A + B)x² + (4A - B + C)x + (4A - 6B - 3C)

(equating coefficients)

A + B = 1

4A - B + C = 0

4A - 6B - 3C = 0

A + B = 1

C = - 4A + B

4A - 6B - 3(- 4A + B) = 0 → 4A - 6B + 12A - 3B = 0

B = - A + 1

C = - 4A + (- A + 1) = - 4A - A + 1 = - 5A + 1

16A - 9(- A + 1) = 0 → 16A + 9A - 9 = 0 → 25A = 9

B = - A + 1 = - (9/25) + 1 = (- 9 + 25)/25 = 16/25

C = - 5A + 1 = - 5(9/25) + 1 = - (9/5) + 1 = (- 9 + 5)/5 = - 4/5

A = 9/25

obtaining:

x² /[(x - 3)(x + 2)²] = A/(x - 3) + B/(x + 2) + C/(x + 2)² =

(9/25)/(x - 3) + (16/25)/(x + 2) + (- 4/5)/(x + 2)²

then the integral becomes:

∫ {[(9/25) /(x - 3)] + [(16/25) /(x + 2)] - [(4/5) /(x + 2)²]} dx =

(splitting into three integrals and factoring constants out)

(9/25) ∫ [1 /(x - 3)] dx + (16/25) ∫ [1 /(x + 2)] dx - (4/5) ∫ (x + 2)‾ ² dx =

(being d(x + 2) the same as dx)

(9/25) ln |x - 3| + (16/25) ln |x + 2| - (4/5) ∫ (x + 2)‾ ² d(x + 2) =

(9/25) ln |x - 3| + (16/25) ln |x + 2| - (4/5) [1/(- 2+1)] (x + 2)‾ ² ⁺ ¹ + C =

(9/25) ln |x - 3| + (16/25) ln |x + 2| - (4/5) [1/(- 1)] (x + 2)‾ ¹ + C =

(9/25) ln |x - 3| + (16/25) ln |x + 2| - (4/5)[- 1 /(x + 2)] + C =

ending with:

(9/25) ln |x - 3| + (16/25) ln |x + 2| + {4 /[5(x + 2)]} + C

I hope it's helpful

In order to integrate ∫x² / [(x-3)(x+2)²] dx, we first must decompose into partial fractions:

x² / [(x-3)(x+2)²] = A/(x-3) + B/(x+2) + C/(x+2)²

Multiply the equation by (x-3)(x+2)²:

x² = A(x+2)² + B(x-3)(x+2) + C(x-3)

When x = -2, we see that 4 = -5C --> C = -4/5

When x = 3, we see that 9 = 25A --> A = 9/25

To find B, we have no more convenient x values to plug in, so we can use a method of comparison terms. For example, the quadratic terms on the right side of the equation are made up of Ax²+Bx², and this must equal x² from the left hand side. Thus, (A+B) = 1 --> 9/25 + B = 1 --> B = 16/25.

So we have the following integral broken up into three pieces, with a simple u-substitution for u = x+2 for the last of the three:

∫x² / [(x-3)(x+2)²] dx = (9/25) ∫dx/(x-3) + (16/25) ∫dx/(x+2) - (4/5) ∫dx/(x+2)²

= (9/25)ln|x-3| + (16/25)ln|x+2| + 4/[5(x+2)] + C