APEX Calculus

The denominator is already factored, as both \(x^2+x+2\) and \(x^2+x+7\) cannot be factored further. We need to decompose \(f(x)\) properly. Since \((x+5)\) is a linear term that divides the denominator, there will be a

term in the decomposition.

As \((x-2)^3\) divides the denominator, we will have the following terms in the decomposition:

The \(x^2+x+2\) term in the denominator results in a \(\ds\frac{Ex+F}{x^2+x+2}\) term.

Finally, the \((x^2+x+7)^2\) term results in the terms

All together, we have

Solving for the coefficients \(A\text{,}\) \(B \ldots J\) would be a bit tedious but not “hard.”

The denominator factors into two linear terms: \(x^2-1 = (x-1)(x+1)\text{.}\) Thus

To solve for \(A\) and \(B\text{,}\) first multiply through by \(x^2-1 = (x-1)(x+1)\text{:}\)

by collecting like terms.

The next step is key. Note the equality we have:

For clarity's sake, rewrite the left hand side as

On the left, the coefficient of the \(x\) term is 0; on the right, it is \((A+B)\text{.}\) Since both sides are equal, we must have that \(0=A+B\text{.}\)

Likewise, on the left, we have a constant term of 1; on the right, the constant term is \((A-B)\text{.}\) Therefore we have \(1=A-B\text{.}\)

We have two linear equations with two unknowns. This one is easy to solve by hand, leading to

If we add these two equations, we get \(2A=1 \Rightarrow A=1/2\text{.}\) Substitution into the first equation gives \(B=-1/2\text{.}\)

Thus

As we saw in Example 6.5.4,

To solve for \(A\) and \(B\) using the Heaviside method, we will build to a common denominator:

Now since the denomiators match, we will only consider the numerator equation (essentially if we multiply both sides of the equation by \((x-1)(x+1)\text{,}\) we will clear the denomiators):

Now we substitute in “convenient” values of \(x\text{.}\) When \(x=1\text{,}\) we get \(1=2A \Rightarrow A=1/2\text{.}\) When \(x=-1\text{,}\) we get \(1=-2B \Rightarrow B=-1/2\text{.}\)

You may note that \(x=1\) and \(x=-1\) were not in the domain of the original fraction. However,

is an identity, meaning it is true for all values of \(x\text{,}\) even those for which the equation is undefined. We could have chosen any values of \(x\) to substitute. Whenever possible, we choose values of \(x\) that will make one of the factors zero. In this way, we can avoid solving a system of equations.

Thus as in Example 6.5.3, we get

We decompose the integrand as follows, as described by Key Idea 6.5.2:

To solve for \(A\text{,}\) \(B\) and \(C\text{,}\) we multiply both sides by \((x-1)(x+2)^2\text{:}\)

Now we collect like terms:

We have

leading to the equations

These three equations of three unknowns lead to a unique solution:

Thus

Each can be integrated with a simple substitution with \(u=x-1\) or \(u=x+2\) (or by directly applying Key Idea 6.1.5 as the denominators are linear functions). The end result is

Key Idea 6.5.2 presumes that the degree of the numerator is less than the degree of the denominator. Since this is not the case here, we begin by using polynomial division to reduce the degree of the numerator. We omit the steps, but encourage the reader to verify that

Using Key Idea 6.5.2, we can rewrite the new rational function as:

for appropriate values of \(A\) and \(B\text{.}\) Clearing denominators, we have

We can now integrate.

The degree of the numerator is less than the degree of the denominator so we begin by applying Key Idea 6.5.2. We have:

Now, letting \(x=-1\) we have \(30 = 6A \Rightarrow A=5\text{.}\) When \(x=0\text{,}\) \(54 = 11A+C\text{.}\) But we know that \(A=1\text{,}\) so \(54 =55+C \Rightarrow C=-1\) Finally, we choose \(x=1\) (with \(A=5, C=-1\)) we have \(92=90+(B-1)(2)\Rightarrow B=2\text{.}\)

Thus

The first term of this new integrand is easy to evaluate; it leads to a \(5\ln\abs{x+1}\) term. The second term is not hard, but takes several steps and uses substitution techniques.

The integrand \(\ds \frac{2x-1}{x^2+6x+11}\) has a quadratic in the denominator and a linear term in the numerator. This leads us to try substitution. Let \(u = x^2+6x+11\text{,}\) so \(du = (2x+6)\, dx\text{.}\) The numerator is \(2x-1\text{,}\) not \(2x+6\text{,}\) but we can get a \(2x+6\) term in the numerator by adding 0 in the form of “\(7-7\text{.}\)”

We can now integrate the first term with substitution, leading to a \(\ln\abs{x^2+6x+11}\) term. The final term can be integrated using arctangent. (We can tell there is no further factoring for this quadratic since the denominator has no real solutions). First, complete the square in the denominator:

An antiderivative of the latter term can be found using Theorem 6.1.20 and substitution:

Let's start at the beginning and put all of the steps together.

As with many other problems in calculus, it is important to remember that one is not expected to “see” the final answer immediately after seeing the problem. Rather, given the initial problem, we break it down into smaller problems that are easier to solve. The final answer is a combination of the answers of the smaller problems.