Integration using partial fractions

[JOhnJDC]

Homework Statement

$$\int$$(3x³-4x²-3x+2)/(x⁴-x²)

Homework Equations

P(x)/Q(x)=A₁/(x-r₁)+A₂/(x-r₂)+...

if x-r occurs with multiplicity m, then A/(x-r) must be replaced by a sum of the form:
B₁/(x-r)+B₂/(x-r)²+...

I think this second equation is the source of my confusion.

The Attempt at a Solution

I began by factoring the denominator:
x⁴-x² = x2(x+1)(x-1)

So, according to my book, we have the following constants:
A/x + B/x² + C/(x+1) + D/(x-1)

First question: where did the x come from in the constant A/x? Does this follow from the rule above? That is, because x² has multiplicity 2, I get A/x and B/x²?

Next, according to my book, when you clear the fractions, you should get:
Ax(x+1)(x-1) + B(x+1)(x-1) + Cx²(x-1) + Dx²(x+1)

I don't understand this. Why isn't it Ax²(x+1)(x-1) + Bx(x+1)(x-1) + Cx³(x-1) + Dx³(x+1)?

Can someone explain? Many thanks.

 

[lanedance]

you multiply everything by the original denominator to get the original fraction

try multiplying out below and it should all become clear ;)
$$\frac{(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{D}{x-1})(x^2(x+1)(x-1))<br /> }{x^(x+1)(x-1)<br /> }$$

 

[Mark44]

Homework Statement

$$\int$$(3x³-4x²-3x+2)/(x⁴-x²)

Homework Equations

P(x)/Q(x)=A₁/(x-r₁)+A₂/(x-r₂)+...

if x-r occurs with multiplicity m, then A/(x-r) must be replaced by a sum of the form:
B₁/(x-r)+B₂/(x-r)²+...

I think this second equation is the source of my confusion.

The Attempt at a Solution

I began by factoring the denominator:
x⁴-x² = x2(x+1)(x-1)

So, according to my book, we have the following constants:
A/x + B/x² + C/(x+1) + D/(x-1)

A, B, C, and D are constants, but the expressions above aren't constant.

First question: where did the x come from in the constant A/x? Does this follow from the rule above? That is, because x² has multiplicity 2, I get A/x and B/x²?

Yes. The denominator in factored form has an x² factor, so the multiplicity of this factor is 2. This means that you'll need to have terms A/x and B/x².

Next, according to my book, when you clear the fractions, you should get:
Ax(x+1)(x-1) + B(x+1)(x-1) + Cx²(x-1) + Dx²(x+1)

I don't understand this. Why isn't it Ax²(x+1)(x-1) + Bx(x+1)(x-1) + Cx³(x-1) + Dx³(x+1)?

Can someone explain? Many thanks.

You have this equation, which has to be identically true (true for all values of x other than those that make any denominator 0).

$$\frac{3x^3 - 4x^2 - 3x + 2}{x^2(x^2 - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1} + \frac{D}{x - 1}$$

You can clear the fractions by multiplying both sides of this equation by x²(x² - 1). When you multiply the terms on the right hand side above, there will be some cancellation.

 

[JOhnJDC]

That clears it up. Thanks, guys.