Math 2411-Notes for February 3, 2004

Integration by Partial Fractions:

Before we start on the new material, consider the following three formulas for antiderivatives:

$$(1)\;\displaystyle \int\displaystyle \frac{du}{u}=\ln\vert u\vert... ...;\;\;\; (3)\;\displaystyle \int u^{n}\;du=\displaystyle \frac{u^{n+1}}{n+1}+\;C$$

In each of the above formulas we assume that $u$ is some function of $x$ and $du$ is the derivative of $u$ with respect to $x$. In formula (3) we assume that $n\not=-1$ (since this would then be formula (1)). A couple of examples will be instructive.

First,

$$\displaystyle \int\displaystyle \frac{2x+1}{x^2+x+3}\;dx=\ln\vert x^2+x+3\vert+C$$

This is an example of formula (1). Formula (1) tells us that if the numerator is the derivative of the denominator then the antiderivtaive is the natural logarithm of the denomimator. Since the derivative of the denominator $d/dx[x^2+x+3]=2x+1$ is the numerator, we fit this formula.

The next example is

$$\displaystyle \int\displaystyle \frac{1}{4+x^2}\;dx= \displaystyle \frac{1}{2}\arctan\left(\displaystyle \frac{x}{2}\right)+C$$

Here we have used formula (2) where $a^2=4\longrightarrow a=2$ and $u^2=x^2\longrightarrow u=x$ and so $du=dx$. Some of you might be tempted to give the solution using formula (1). But the derivative of the denominator is $d/dx[4+x^2]=2x$, and the numerator is not the derivative of the denominator. Formula (1) does not help. We should consider one more example of this $\arctan$ ``form.''

$$\displaystyle \int\displaystyle \frac{2x}{3+4x^4}\;dx$$

Again it is tempting to write the solution using formula (1). But the derivative of the denominator is $d/dx[3+4x^4]=16x^3$ which does not match the numerator. So we see if it fits formula (2). We would have $a^2=3\longrightarrow a=\sqrt{3}$, and $u^2=4x^4\longrightarrow u=2x^2$. Then since $2x\;dx=(4x\;dx)/2=du/2$ we can rewrite the integral as

$$\displaystyle \int\displaystyle \frac{2x}{3+4x^4}\;dx=\displaystyle \frac{1}{2}\displaystyle \int\displaystyle \frac{du}{a^2+u^2}.$$

So the solution is

$$\displaystyle \int\displaystyle \frac{2x}{3+4x^4}\;dx = \displaystyle \frac{1}{2}\cdot\left[\displaystyle \frac{1}{\sqrt{3}} \arctan\left(\displaystyle \frac{2x^2}{\sqrt{3}}\right)\right]+C$$

$$= \displaystyle \frac{\sqrt{3}}{6}\arctan\left(\displaystyle \frac{2\sqrt{3}}{3}\cdot x^2\right)+C$$

Now for an example of formula (3).

$$\displaystyle \int\displaystyle \frac{1}{(x+1)^2}\;dx=-\displaystyle \frac{1}{x+1}+C$$

To see that this follows formula (3) notice that we can rewrite the integral as

$$\displaystyle \int\displaystyle \frac{1}{(x+1)^2}\;dx=\displaystyle \int(x+1)^{-2}\;dx$$

Here $u=x+1$ and we use the power rule of formula (3).

* Notice that we used the familiar technique $u$-substitution in all three of these formulas. In other words, we were simply seeing if these inetgrals fit one of the patterns in these formulas. Lets reiterate the important ideas in each formula:

$\bullet$ When integrating a rational function (a fraction) and the numerator is the derivative of the denominator our solution will involve the natural logarithm of the denominator.

$\bullet$ If the denominator is some function raised to a power (besides 1) and the numerator is the derivative of the inside piece of the denominator, then we use the power rule in formula (3).

$\bullet$ If the denominator is the sum of a constant and some function of $x$, and the numerator is the derivative of the square root of that function of $x$, then we use the formula (2).

Now that we see how these formulas work we should ask how to deal with rational functions that do not match any of the patterns in these three formulas we just visited. Consider

$$\displaystyle \int\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}\;dx$$

We should first notice that formula (1) does not work since the derivative of the denominator $d/dx[x^3-3x^2+x-3]=3x^2-6x+1$ does not match the numerator. The denominator is not some polynomial raised to an exponent power and is clearly not some constant added to a function squared. If we use our calculators we arrive at the solution

$$\displaystyle \int\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}\;dx= \displaystyle \frac{1}{2}\ln(x^2+1)-\ln\vert x-3\vert+\arctan(x)+C$$

If we take derivatives of both sides we should arrive at some identity equation for the integrand.

$$\displaystyle \frac{d}{dx}\displaystyle \int\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}\;dx = \displaystyle \frac{d}{dx}\left[\displaystyle \frac{1}{2}\ln(x^2+1)-\ln\vert x-3\vert+\arctan(x)+C\right]$$

$$\downarrow$$

$$\displaystyle \frac{-2x-4}{x^3-3x^2+x-3} = \displaystyle \frac{x}{x^2+1}-\displaystyle \frac{1}{x-3}+\displaystyle \frac{1}{1+x^2}$$

$$= \displaystyle \frac{x+1}{x^2+1}-\displaystyle \frac{1}{x-3}$$

It appears that to solve this problem we need to break the initial fraction into a number of simpler pieces. The technique that we use to rewrite a fraction as the sum of a number of simpler fractions is called partial fractions.

To perform a partial fraction decomposition of a rational function we follow a number of easy steps. Lets follow the steps with using our current example.

1. Completely factor the denominator into irreducible factors. Recall that every polynomial can be factored into a product of linear and irreducible quadratic factors. (A quadratic factor is irreduceable if it has no real roots, or zeros.)
   $$x^3-3x^2+x-3=(x^2+1)(x-3)$$

2. Each factor of the denominator will give us a piece of the decomposition, where the numerator over a quadratic factor is an arbitary linear function, and the numerator of a piece corresponding to a linear factor will be an arbitrary constant.
   $$\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}=\displaystyle \frac{-2x-4}{(x^2+1)(x-3)}= \displaystyle \frac{Ax+B}{x^2+1}+\displaystyle \frac{C}{x-3}$$

3. We are now ready to solve for the constants in the numerators. We do this by first combining all the new ``pieces'' into a single fraction which has a denominator that matches the original denominator.

   
   

   $$\displaystyle \frac{-2x-4}{x^3-3x^2+x-3} = \displaystyle \frac{Ax+B}{x^2+1}+\displaystyle \frac{C}{x-3}$$

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

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

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

   
   

4. Now that we have completed the recombination notice that we have two equal fractions, both having the same denominator. So their numerators must also match.
   $$-2x-4=(A+C)x^2+(B-3A)x+(C-3B)$$

   It follows (by matching the coefficients of these polynomials) that
   

   $$A+C =$$ 0

   $$B-3A = -2$$

   $$C-3B = -4$$

   
   

   This is now a system of 3 linear equations with three variables. We get
   

   $$A+C=0 \longrightarrow A=-C$$ (1)

   $$B-3A=-2 \longrightarrow B=3A-2$$ (2)

   $$C-3B=-4 \longrightarrow C=3B-4$$ (3)

   $$equation\;(2) \longrightarrow C=3(3A-2)-4$$ (4)

   $$\longrightarrow C=9A-10$$ (5)

   $$equation\;(1) \longrightarrow C=9(-C)-10$$ (6)

   $$\longrightarrow C=-1$$ (7)

   $$equation\;(1) \longrightarrow A=1$$ (8)

   $$equation\;(2) \longrightarrow B=3(1)-2=1$$ (9)

   
   

5. Now we substitute these variables and integrate to get
   

   $$\displaystyle \frac{-2x-4}{x^3-3x^2+x-3} = \displaystyle \frac{Ax+B}{x^2+1}+\displaystyle \frac{C}{x-3}$$

   $$= \displaystyle \frac{x+1}{x^2+1}-\displaystyle \frac{1}{x-3}$$

   $$\downarrow$$

   $$\displaystyle \int\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}\;dx = \displaystyle \int\displaystyle \frac{x+1}{x^2+1}-\displaystyle \frac{1}{x-3}\;dx$$

   $$= \displaystyle \int\displaystyle \frac{x+1}{x^2+1}\;dx-\displaystyle \int\displaystyle \frac{1}{x-3}\;dx$$

   
   

   Its time to go back to the formulas from the begining of the notes. The second integral is easiest since the numerator is the derivative of the denominator; i.e., let $u=x-3$ and then $du=dx$ and we rewrite the integral as

   $$\displaystyle \int\displaystyle \frac{1}{x-3}\;dx\longrightarrow\displaystyle \int\displaystyle \frac{du}{u}=\ln\vert u\vert+C$$
   Hence
   $$\displaystyle \int\displaystyle \frac{1}{x-3}\;dx=\ln\vert x-3\vert+C$$

   What about the next piece. Check that we don't fit into any of the three formulas. Furthermore, we can not perform partial fractions again because the denominator can not be factored. We might seem to be stuck, but don't forget that one can always use basic algebra and break up the numerator.

   
   

   $$\displaystyle \int\displaystyle \frac{x+1}{x^2+1}\;dx = \displaystyle \int\displaystyle \frac{x}{x^2+1}+\displaystyle \frac{1}{x^2+1}\;dx$$

   $$= \displaystyle \int\displaystyle \frac{x}{x^2+1}\;dx+\displaystyle \int\displaystyle \frac{1}{x^2+1}\;dx$$

   
   

   Now the first piece is solved using $u$-substitution and letting $u=x^2+1$. Then $du=2x\;dx$ and the numerator is (a scalar multiple of) the derivative of the denominator and we use formula (1). The second piece fits into formula (2) with $a^2=1\;\longrightarrow a=1$ and $u^2=x^2\;\longrightarrow u=x$. So we get

   $$\displaystyle \int\displaystyle \frac{x}{x^2+1}\;dx+\displaystyle... ...playstyle \frac{1}{x^2+1}\;dx= \displaystyle \frac{1}{2}\ln(x^2+1)+\arctan(x)+C$$
   Combining all the information we get

   
   

   $$\displaystyle \frac{d}{dx}\displaystyle \int\displaystyle \frac{-2x-4}{x^3-3x^2+x-3}\;dx = \displaystyle \int\displaystyle \frac{x+1}{x^2+1}\;dx-\displaystyle \int\displaystyle \frac{1}{x-3}\;dx$$

   $$= \displaystyle \frac{1}{2}\ln(x^2+1)+\arctan(x)-\ln\vert x-3\vert+C$$

   
   
   which is exactly the solution we found with technology.
   

Lets write out the basic formula ued in the initial decomposition in step 2 above.

$$\displaystyle \frac{p(x)}{(ax+b)^k(cx^2+dx+e)^m}$$ *

$$= \displaystyle \frac{A}{ax+b}+ \displaystyle \frac{B}{(ax+b)^2}+\cdots+\displaystyle \frac{D}{(ax+b)^k}+$$

$$\displaystyle \frac{Ex+F}{cx^2+dx+e}+\displaystyle \frac{Gx+H}{(cx^2+dx+e)^2}+ \cdots+\displaystyle \frac{Kx+L}{(cx^2+dx+e)^m}$$

Notice that the general formula accounts for repeated roots which did not appear in our example. As an example study

$$\displaystyle \frac{p(x)}{x^2(x-1)(x+2)^3(7x^2+1)^2} = \displaystyle \frac{A}{x-1}+ \displaystyle \frac{B}{x+2}+\displaystyle \frac{C}{(x+2)^2}+\displaystyle \frac{D}{(x+2)^3}+$$

$$\displaystyle \frac{E}{x}+\displaystyle \frac{F}{x^2}+\displaystyle \frac{Gx+H}{7x^2+1}+ \displaystyle \frac{Kx+L}{(7x^2+1)^2}$$

The only tricky part of this example is noticing that $x^2$ is a repeated linear factor because it has real roots; i.e., $x^2$ can still be factored into $x\cdot x$ which is the same as $(x-0)(x-0)=(x-0)^2$ which is a repated linear factor that has the root zero.

Of course it would be a nightmare to determine all the constants $A,B,C,D,E,F,G,H,K,L$ by hand, but this examples illustrates well the genral technique in step 2.