Integration - Partial Fraction Decomposition

• Mar 19th 2010, 06:44 PM
VitaX
Integration - Partial Fraction Decomposition
$\int \frac{x^2 + 3x + 1}{x^4 + 5x^2 + 4}dx$

So first step would be factoring the denominator. $x^4 + 5x^2 + 4 = (x^2 + 1)(x^2 + 4)$

$\int \frac{x^2 + 3x + 1}{(x^2 + 1)(x^2 + 4)}dx$

$\frac{A}{(x^2 + 1)} + \frac{Bx + C}{(x^2 + 4)}$

$x^2 + 3x + 1 = A(x^2 + 4) + (Bx + C)(x^2 + 1)$

Do I have this set up correctly so far?

If so next step would be to find A, so you let x = 0 I believe.

$0^2 + 3(0) + 1 = A(0^2 + 4) \rightarrow 1 = 4A \rightarrow A = \frac{1}{4}$

To find C, let x = 0 $0^2 + 3(0) + 1 = 4A + (B(0) + C)(0^2 +1) \rightarrow 1 = 4\left(\frac{1}{4}\right) + C \rightarrow C = 1$

If this is correct so far how would I find B, if this is not correct can someone guide me along here?
• Mar 19th 2010, 07:23 PM
Prove It
Quote:

Originally Posted by VitaX
$\int \frac{x^2 + 3x + 1}{x^4 + 5x^2 + 4}dx$

So first step would be factoring the denominator. $x^4 + 5x^2 + 4 = (x^2 + 1)(x^2 + 4)$

$\int \frac{x^2 + 3x + 1}{(x^2 + 1)(x^2 + 4)}dx$

$\frac{A}{(x^2 + 1)} + \frac{Bx + C}{(x^2 + 4)}$

$x^2 + 3x + 1 = A(x^2 + 4) + (Bx + C)(x^2 + 1)$

Do I have this set up correctly so far?

If so next step would be to find A, so you let x = 0 I believe.

$0^2 + 3(0) + 1 = A(0^2 + 4) \rightarrow 1 = 4A \rightarrow A = \frac{1}{4}$

To find C, let x = 0 $0^2 + 3(0) + 1 = 4A + (B(0) + C)(0^2 +1) \rightarrow 1 = 4\left(\frac{1}{4}\right) + C \rightarrow C = 1$

If this is correct so far how would I find B, if this is not correct can someone guide me along here?

Actually you need to do

$\frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{x^2 + 1} = \frac{x^2 + 3x + 1}{(x^2 + 4)(x^2 + 1)}$.

This means

$(Ax + B)(x^2 + 1) + (Cx + D)(x^2 + 4) = x^2 + 3x + 1$

$Ax^3 + Ax + Bx^2 + B + Cx^3 + 4Cx + Dx^2 + 4D = x^2 + 3x + 1$

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x + B + 4D = x^2 + 3x + 1$.

Equating coefficients of like powers of $x$ gives:

$A + C = 0$

$B + D = 1$

$A + 4C = 3$

$B + 4D = 1$.

Solving these equations simultaneously gives:

$A = -1, B = 1, C = 1, D = 0$.

So that means:

$\frac{x^2 + 3x + 1}{(x^2 + 4)(x^2 + 1)} = \frac{1 - x}{x^2 + 4} + \frac{x}{x^2 + 1}$.

Now when you integrate, you will need to use trigonometric substitution on the first term and $u$ substitution on the second.
• Mar 19th 2010, 07:24 PM
skeeter
Quote:

Originally Posted by VitaX
$\int \frac{x^2 + 3x + 1}{x^4 + 5x^2 + 4}dx$

So first step would be factoring the denominator. $x^4 + 5x^2 + 4 = (x^2 + 1)(x^2 + 4)$

$\int \frac{x^2 + 3x + 1}{(x^2 + 1)(x^2 + 4)}dx$

$\frac{x^2 + 3x + 1}{(x^2 + 1)(x^2 + 4)}=\textcolor{red}{\frac{Ax+B}{(x^2 + 1)} + \frac{Cx + D}{(x^2 + 4)}}$

...
• Mar 19th 2010, 07:34 PM
VitaX
Quote:

Originally Posted by Prove It
Actually you need to do

$\frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{x^2 + 1} = \frac{x^2 + 3x + 1}{(x^2 + 4)(x^2 + 1)}$.

This means

$(Ax + B)(x^2 + 1) + (Cx + D)(x^2 + 4) = x^2 + 3x + 1$

$Ax^3 + Ax + Bx^2 + B + Cx^3 + 4Cx + Dx^2 + 4D = x^2 + 3x + 1$

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x + B + 4D = x^2 + 3x + 1$.

Equating coefficients of like powers of $x$ gives:

$A + C = 0$

$B + D = 1$

$A + 4C = 3$

$B + 4D = 1$.

Solving these equations simultaneously gives:

$A = -1, B = 1, C = 1, D = 0$.

So that means:

$\frac{x^2 + 3x + 1}{(x^2 + 4)(x^2 + 1)} = \frac{1 - x}{x^2 + 4} + \frac{x}{x^2 + 1}$.

Now when you integrate, you will need to use trigonometric substitution on the first term and $u$ substitution on the second.

My main question is how do you know how to set it up. Why is it Ax + b + Cx + d exactly? I guess I should have taken Pre-Calc (Speechless)
• Mar 19th 2010, 07:38 PM
Prove It
Quote:

Originally Posted by VitaX
My main question is how do you know how to set it up. Why is it Ax + b + Cx + d exactly? I guess I should have taken Pre-Calc (Speechless)

Your numerators have to have a degree smaller than the denominator, but you also have to write the highest possible degree numerator.

In other words, the numerator is always written as the polynomial of one less degree than the denominator. Since you have quadratic functions in your denominators, the numerators can be anything up to a linear function.
• Mar 19th 2010, 07:43 PM
VitaX
Quote:

Originally Posted by Prove It
Your numerators have to have a degree smaller than the denominator, but you also have to write the highest possible degree numerator.

In other words, the numerator is always written as the polynomial of one less degree than the denominator.

Ok so since the highest degree was 4 we have 4 terms in the numerator makes sense. But in my notes why is it that sometimes you write them as A + B + C and other times its A + Bx + C. My last question would be on the part where you wrote "equating coefficients of like powers of x". What exactly are you doing there to get those answers? I must not be seeing something clearly again.
• Mar 19th 2010, 07:49 PM
Prove It
Quote:

Originally Posted by VitaX
Ok so since the highest degree was 4 we have 4 terms in the numerator makes sense. But in my notes why is it that sometimes you write them as A + B + C and other times its A + Bx + C. My last question would be on the part where you wrote "equating coefficients of like powers of x". What exactly are you doing there to get those answers? I must not be seeing something clearly again.

No, the degree has nothing to do with how many terms.

Remember that the degree of a polynomial is the highest power of $x$.
Since you reduced it down to fractions with quadratic functions (degree 2) in the denominators, that means that the numerators could be anything up to a linear function (degree 1).

You have

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x + B + 4D = x^2 + 3x + 1$.

Notice that I could rewrite the the equation so that it looks like this:

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x^1 + (B + 4D)x^0 = 0x^3 + 1x^2 + 3x^1 + 1x^0$.

The left hand side can only equal the right hand side if the coefficients of like powers of $x$ are equal.

Since the coefficient of $x^3$ is $0$, that means for LHS to equal RHS, $A + C = 0$.

Similarly, the coefficient of $x^2$ is $1$. So for LHS to equal RHS, $B + D = 1$.

Using the same logic, $A + 4C = 3$ and $B + 4D = 1$.

That gives you four equations in four unknowns that you can solve simultaneously.
• Mar 19th 2010, 07:59 PM
VitaX
Quote:

Originally Posted by Prove It
No, the degree has nothing to do with how many terms.

Remember that the degree of a polynomial is the highest power of $x$.
Since you reduced it down to fractions with quadratic functions (degree 2) in the denominators, that means that the numerators could be anything up to a linear function (degree 1).

You have

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x + B + 4D = x^2 + 3x + 1$.

Notice that I could rewrite the the equation so that it looks like this:

$(A + C)x^3 + (B + D)x^2 + (A + 4C)x^1 + (B + 4D)x^0 = 0x^3 + 1x^2 + 3x^1 + 1x^0$.

The left hand side can only equal the right hand side if the coefficients of like powers of $x$ are equal.

Since the coefficient of $x^3$ is $0$, that means for LHS to equal RHS, $A + C = 0$.

Similarly, the coefficient of $x^2$ is $1$. So for LHS to equal RHS, $B + D = 1$.

Using the same logic, $A + 4C = 3$ and $B + 4D = 1$.

That gives you four equations in four unknowns that you can solve simultaneously.

Ok makes a lot more sense now, I guess we just didn't get this deep into PFD in class, never would have known to solve it like this. I'll try and the integral now and post my work when I'm done.
• Mar 19th 2010, 08:17 PM
drumist
Quote:

Originally Posted by Prove It
No, the degree has nothing to do with how many terms.

Actually it does always work out such that the number of constants equals the degree of the original denominator.
• Mar 19th 2010, 08:20 PM
Prove It
Quote:

Originally Posted by drumist
Actually it does always work out such that the number of constants equals the degree of the original denominator.

OK fair enough - I just never think about it (Rofl)
• Mar 19th 2010, 08:25 PM
VitaX
Quote:

Originally Posted by drumist
Actually it does always work out such that the number of constants equals the degree of the original denominator.

Yah I noticed that later on after looking through my notes. Thanks for that.

And also my answer equals wolf rams answer so everything is good. I just really hate PFD. Need more practice in it for sure.
• Mar 20th 2010, 12:51 AM
simplependulum
You got $x^4 + 5x^2 + 4 = (x^2+1)(x^2+4)$

I observe the numerator $x^2 + 3x + 1 = (x^2+1) + 3x$

so the fraction can be expressed as :

$\frac{(x^2+1) + 3x }{ (x^2+1)(x^2+4)}$

$= \frac{1}{x^2+4} + \frac{3x}{(x^2+1)(x^2+4)}$

the integral $=$

$\frac{1}{2}\arctan(\frac{x}{2}) + 3\int \frac{x}{(x^2+1)(x^2+4)} ~dx$

Sub $x^2 = t \implies dx = \frac{dt}{2}$

the integral becomes

$\frac{3}{2} \int \frac{dt}{ (t+1)(t+4) }$ which just requires you to resolve a much simplier fraction .

$\frac{1}{2} \arctan( \frac{x}{2} ) + \frac{1}{2} \ln(x^2 + 1) - \frac{1}{2} \ln(x^2+4) + C$