Hairy Integral by Partial Fractions
A reader asked how to solve this integral.
This kind of integrals that involve rational function are usually solved a technique called partial fractions. This is an algebraic technique that converts the complicated rational function there to the sum of simpler functions. These simpler functions we'll know how to integrate.
The first step in developing the expression in partial fractions is to factor the denominator. This partical example is a particularly difficult one because the denominator is of degree 3. This means it is harder to factor. I'll tell you how I did it, but if you found an easier way to figure it out, please leave a comment below.
If a number "a" is a root of a polynomial, it means that the polynomial has a factor (x-a) in its expression. So, what we must do to factor the denominator is to find a root. First of all, let's give a name to the polynomial and evaluate it at two points:
We have that this function is negative at x=0 and positive at x=1. This means that the function must have crossed the x axis
at a point between x=0 and x=1. This is intuitive, because the graph of the function must do something like this:
Now we need to locate the precise point at which this occurs. This can be done using Newton's method. This is the fastest way I can think of. If you found an easier way, please talk about it in the comments.
If you don't know what Newton's method is and don't want to learn it (it isn't difficult), you may need to trust me that a root of this polynomial is x=0.5. In fact, you may verify it easily.
So, now, we can write this polynomial as the product of (x-0.5) and another polynomial (of degree 2). This another polynomial of degree two is what we need to determine now. This can be done simply multiplying out and using the equalities between the coefficients:
We need to solve for A, B and C here. Using the equalities between the coefficients we have that:
So, our polynomial equals:
A more convenient way of writing this is factoring two out of the second parenthesis and putting into the first parenthesis:
You can check that this equality is sound just by performing the multiplication in the second member. (Another way to do this is to use polynomial division: if you divide the original polynomial by the factor (2x-1) you should get the second factor).
Now, we're ready to apply partial fractions. What we'll try to do is to write the function inside the integral sign as:
Here F, G and H are constants we want to determine. Note that each of the terms in the second member we know how to integrate.
How do I know it can be written in this form? We'll there are some rules to partial fractions. The most important are:
- For factors that have real roots, that is, factors of the form (x-a) use the partial fraction:
- For factors that have only complex roots, use factors of the form:
If you use these simple rules, you'll always be able to expand rational function into partial fractions.
Now, let's return to our example. We know that the expression for our function should be:
What we need to do now is to sum the fractions in the right side and solve for the unkown constants(using the equalities of each coefficient in the numerator:
Using the equalities of each coefficient in the numerator we get three equations:
The solution of this system is:
So, there we have the partial fractions expansion of our function:
Now we only need to take integral of each of those:
And our integral is simply the sum of these:
These problems can be long, but if you apply the rules for partial fractions you'll be okay.Return to Indefinite Integrals