Question

Different methods Let \(I=\int \frac{x+2}{x+4} d x\) a. Evaluate \(I\) after first performing long division on the integrand. b. Evaluate \(I\) without performing long division on the integrand. c. Reconcile the results in parts (a) and (b).

Short answer

Now that we have evaluated the integral using both methods, we can see that both methods yield equivalent results, up to the constant of integration. By performing long division on the integrand or evaluating the integral directly, we obtain the same result, which is: \(I = \int \frac{x+2}{x+4}\ dx = x - 2\ln|x+4| + C\)

Step-by-step solution

Method a: Long division

To perform long division on the integrand, we divide \(x+2\) by \(x+4\): $$ \begin{array}{c|cc} \multicolumn{2}{r}{1} & - \frac{2}{x+4} \\ \cline{2-3} x+4 & x+2 & \\ \cline{2-2} \multicolumn{2}{r}{-2} \\ \cline{2-2} \end{array} $$ Thus, after performing long division, the integrand is: \(\frac{x+2}{x+4} = 1 - \frac{2}{x+4}\) Now we can evaluate the integral: \(I = \int \frac{x+2}{x+4}\ dx = \int (1 - \frac{2}{x+4})\ dx\) Separate the integration: \(I = \int 1\ dx - \int \frac{2}{x+4}\ dx\) \(I = x - 2 \int \frac{1}{x+4}\ dx\) Let \(u = x + 4\), then \(du = dx\): \(I = x - 2 \int \frac{1}{u}\ du\) \(I = x - 2(\ln|u| + C)\) Replace \(u\) with \(x+4\): \(I = x - 2\ln|x+4| + C\)

Method b: Direct integration

Now, let's evaluate the integral without long division. Make a substitution, let \(u = x + 4\), then \(du = dx\): \(I = \int \frac{x+2}{x+4}\ dx = \int \frac{u-2}{u}\ du\) Separate the fraction: \(I = \int (1 - \frac{2}{u})\ du\) Evaluate the integral: \(I = \int 1\ du - \int \frac{2}{u}\ du\) \(I = u - 2 \int \frac{1}{u}\ du\) \(I = u - 2(\ln|u| + C)\) Replace \(u\) with \(x+4\): \(I = (x+4) - 2\ln|x+4| + C\)

Reconciling the results

Now, let's reconcile the results from Method a and Method b: Method a: \(I = x - 2\ln|x+4| + C\) Method b: \(I = (x+4) - 2\ln|x+4| + C\) Since the constants of integration, \(C\), can absorb any constant difference, we can say that the results from both methods are equivalent up to the constant of integration. Thus, the integral evaluates to: \(I = \int \frac{x+2}{x+4}\ dx = x - 2\ln|x+4| + C\)

Key concepts

Long division in integration

Sometimes when integrating rational functions, performing long division on the integrand can simplify the process. Long division helps us when the degree of the numerator is greater than or equal to the degree of the denominator. Let's break it down with an example:
Suppose we have to integrate \(\int \frac{x+2}{x+4} \,dx\). The polynomial in the numerator, \(x+2\), has the same degree as the polynomial in the denominator, \(x+4\). This indicates that long division can be helpful.

• We divide \(x+2\) by \(x+4\), which results in \(1 - \frac{2}{x+4}\).
• This means the original integrand is transformed into a simpler form: \(1 - \frac{2}{x+4}\).

Now, evaluating the integral is straightforward because we have two simpler integrals:

• \(\int 1 \,dx = x\)
• \(\int \frac{2}{x+4} \,dx = 2\ln|x+4|\)

Thus, the integral using long division becomes \(x - 2\ln|x+4| + C\). Long division helps in breaking down a complex integrand into more approachable parts.

Substitution in integration

Another powerful tool in integration is substitution, which can simplify integrals by changing the variable. Substitution is particularly useful when dealing with compositions of functions. In our example, direct substitution can also be useful.
Consider \(\int \frac{x+2}{x+4} \,dx\). Here, choosing a substitution for the denominator helps simplify the integrand:

• Let \(u = x + 4\), hence \(du = dx\).
• The original integrand \(\frac{x+2}{x+4}\) can be rewritten, by expressing \(x+2\) as \((u-2)\).

After substitution, the integrand becomes \(\frac{u-2}{u}\) or \(1 - \frac{2}{u}\), and the integral is now in terms of \(u\):

• \(\int 1 \,du = u\)
• \(\int \frac{2}{u} \,du = 2\ln|u|\)

After evaluating these simpler integrals, substitute back \(u = x + 4\):
The result is \((x+4) - 2\ln|x+4| + C\), which is once again a simplified representation of the original integral.

Integral of rational functions

Integration often involves rational functions, which are expressions of the form \(\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials. Solving these integrals can require specific techniques such as long division or substitution.
For instance, consider \(\int \frac{x+2}{x+4} \,dx\). This rational function consists of a numerator and a denominator, both polynomials. Depending on their degrees, different strategies might be employed:

• **Long division:** When the degree of the numerator is equal to or greater than the denominator, simplify by dividing first.
• **Substitution:** Look for substitutions that can make the fraction easier to integrate.

In cases where neither a straightforward substitution nor division is evident, partial fraction decomposition might be another strategy, but that's beyond this scope.
Ultimately, the integral of rational functions requires understanding which method or combination of methods will lend itself to simplifying the problem to a standard form, often resulting in functions involving logarithms when the integral is performed.