How to find these partial fractions

1. (2x-3)/(x-1)^2
Since it's square, how would i find both values of A and B?

2. integral of (6x)/(x^3-8)
so when I factor it becomes (x-2)(x^2+2x+4) would I continue on like this and then but Bx+C in my numerator?

3. (x^2-x+9)/(x^2+9)^2
Again, because it's squared how can I find each separate value of the numerator?

Thanks in advance!

1.

(2x − 3) / (x − 1)² = A/(x − 1) + B/(x − 1)²

Multiply both sides by (x − 1)²

2x − 3 = A (x − 1) + B

Let x = 1, and solve for B:
2 − 3 = 0A + B -----> B = −1

Let x = 0, and solve for A:
0 - 3 = −A + B
−3 = −A − 1
A = 2

(2x − 3) / (x − 1)² = 2/(x − 1) − 1/(x − 1)²

______________________________


2.

Yes, use Bx + C in numerator for denominator x² + 2x + 4

6x / (x³ − 8) = A/(x − 2) + (Bx + C) / (x² + 2x + 4)

Multiply both sides by (x − 2) (x² + 2x + 4)

6x = A (x² + 2x + 4) + (Bx + C) (x − 2)

Let x = 2, and solve for A
12 = A(4 + 4 + 4) + 0
A = 1

Let x = 0 and solve for C
0 = 4A + 0B - 2C
2C = 4A
2C = 4
C = 2

Let x = 1 and solve for B
6 = A (1 + 2 + 4) + (B + C) (−1)
6 = 7A − B − C
B = 7(1) − 2 − 6
B = −1

6x / (x³ − 8) = 1/(x − 2) + (−x + 2) / (x² + 2x + 4)

------------------------------

Before you can integrate, you need to split second fraction up, so that one fraction has numerator that is a multiple of derivative of denominator, and other fraction has constant in numerator.

6x / (x³ − 8) = 1/(x − 2) + ((−x − 1) + 3) / (x² + 2x + 4)
6x / (x³ − 8) = 1/(x − 2) − 1/2 (2x + 2)/(x² + 2x + 4) + 3/(x² + 2x + 4)

Now you complete the square in denominator:

6x / (x³ − 8) = 1/(x − 2) − 1/2 (2x + 2)/(x² + 2x + 4) + 3/((x + 1)² + 3)
6x / (x³ − 8) = 1/(x − 2) − 1/2 (2x + 2)/(x² + 2x + 4) + 1/[((x + 1)/3)² + 1]

∫ 6x / (x³ − 8) dx
= ∫ 1/(x − 2) dx − 1/2 ∫ (2x + 2)/(x² + 2x + 4) dx + ∫ 1/[((x + 1)/√3)² + 1] dx
= ln |x − 2| − 1/2 ln |x² + 2x + 4| + √3 arctan((x + 1)/√3)² + C

______________________________


3.

(x² − x + 9) / (x² + 9)² = (Ax + B)/(x² + 9) + (Cx + D)/(x² + 9)²

Multiply both sides by (x² + 9)²

x² − x + 9 = (Ax + B) (x² + 9) + Cx + D

Expand right side

x² − x + 9 = Ax³ + Bx² + 9Ax + 9B + Cx + D
x² − x + 9 = Ax³ + Bx² + (9A + C) x + (9B + D)
0x³ + x² − x + 9 = Ax³ + Bx² + (9A + C) x + (9B + D)

Matching coefficients we get
0x³ = Ax³ -----> A = 0
x² = Bx² -----> B = 1
−x = (9A+C)x -----> 9(0) + C = −1 -----> C = −1
9 = 9B+D -----> 9(1)+ D = 9 -----> D = 0

(x² − x + 9) / (x² + 9)² = (0x + 1)/(x² + 9) + (−1x + 0)/(x² + 9)²

(x² − x + 9) / (x² + 9)² = 1/(x² + 9) − x/(x² + 9)²

Mαthmφm