1. ## Basic Algebra?

Problem
$\int\frac{10x-2x^2}{(x-1)^2(x+3)}$
I did all the steps correctly until this part

A+C = -2
2A+ B -2C = 10
-3A +3B + C = 0

Attempt:
I said C = -A-2
B = -2A +2C + 10

So
-3A + 3(2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(2A -2A -4 + 10) -A-2 = 0
-3A + -12 +30 -A - 2 = 0
-4A +16 = 0
-4A = -16

A = 4

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.

2. Hi

So
-3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(-2A -2A -4 + 10) -A-2 = 0
-3A + -12A -12 +30 -A - 2 = 0
-16A +16 = 0
-16A = -16

A = 1

3. Originally Posted by Fallen186
Problem
$\int\frac{10x-2x^2}{(x-1)^2(x+3)}$
I did all the steps correctly until this part

A+C = -2
2A+ B -2C = 10
-3A +3B + C = 0

Attempt:
I said C = -A-2
B = -2A +2C + 10

So
-3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(-2A -2A -4 + 10) -A-2 = 0
-3A + -12 +30 -A - 2 = 0
-4A +16 = 0
-4A = -16

A = 4

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.
You missed a sign at the beginning (See chage in red). See if that fixes it. I didn't bother to work it out for you.

4. Originally Posted by Fallen186
Problem
$\int\frac{10x-2x^2}{(x-1)^2(x+3)}$

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.

$\frac{10x-2x^2}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3}
$

$10x-2x^2 = A(x-1)(x+3) + B(x+3) + C(x-1)^2$

$x = 1$ ... $4B = 8$ ... $B = 2$

$x = -3$ ... $16C = -48$ ... $C = -3$

$x = 0$ ... $-3A + 3B + C = 0$ ... $-3A + 6 - 3 = 0$ ... $A = 1$