Question

Evaluate the given integral. $$ \int_{1}^{2} \frac{1}{x^{2}+6 x+8} d x $$

Short answer

\( \frac{1}{2} \ln \frac{10}{9} \)

Step-by-step solution

Factor the denominator

First, notice that the denominator of the integrand is a quadratic expression. We need to factor it to simplify the integrand. The expression is \(x^2 + 6x + 8\). We look for two numbers that multiply to 8 and add to 6. These numbers are 2 and 4. Thus, we can write: \((x^2 + 6x + 8) = (x+2)(x+4)\).

Set up partial fraction decomposition

Next, express the integrand as a sum of partial fractions. Assume: \[\frac{1}{(x+2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4}\]Multiply through by the denominator \(x^2 + 6x + 8\) to clear the fractions: \[1 = A(x+4) + B(x+2)\].

Solve for A and B

Expand and simplify the equation from the partial fractions: \\[1 = Ax + 4A + Bx + 2B = (A+B)x + (4A + 2B)\]\For the equation to hold for all x, the coefficients must match: \- \(A + B = 0\) for the x terms- \(4A + 2B = 1\) for the constant terms.Solving these linear equations, we first find from \(A + B = 0\) that \(B = -A\). Substituting \(B = -A\) into \(4A + 2B = 1\), gives \(2A = 1\), hence \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).

Write the integral with partial fractions

Substitute A and B back into the partial fractions: \[\frac{1}{(x+2)(x+4)} = \frac{1/2}{x+2} - \frac{1/2}{x+4}. \] Now, substitute back into the original integral: \[\int_{1}^{2} \left( \frac{1/2}{x+2} - \frac{1/2}{x+4} \right) dx. \] This separates into two integrals: \[\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} dx - \frac{1}{2} \int_{1}^{2} \frac{1}{x+4} dx. \]

Integrate each term separately

Integrate each part separately using the formula \(\int \frac{1}{x} dx = \ln|x| + C\):\- \(\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} dx = \frac{1}{2} \left[ \ln|x+2| \right]_{1}^{2}\).- \(\frac{1}{2} \int_{1}^{2} \frac{1}{x+4} dx = \frac{1}{2} \left[ \ln|x+4| \right]_{1}^{2}\).Evaluate these results to find:- \(\frac{1}{2}(\ln 4 - \ln 3)\) for the first integral,- \(\frac{1}{2}(\ln 6 - \ln 5)\) for the second.

Combine the results and simplify

Combine the results of the two integrals:\[\frac{1}{2}(\ln 4 - \ln 3) - \frac{1}{2}(\ln 6 - \ln 5) = \frac{1}{2}(\ln \frac{4}{3} - \ln \frac{6}{5}).\]This can be written as:\[\frac{1}{2} \ln \left(\frac{4}{3} \times \frac{5}{6}\right) = \frac{1}{2} \ln \frac{20}{18} = \frac{1}{2} \ln \frac{10}{9}.\]

Final answer

The evaluated integral is: \[\frac{1}{2} \ln \frac{10}{9}.\] This is the final simplified form of the result.

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex fractional expressions into simpler, easier-to-integrate parts. This method is especially useful when dealing with rational functions where the degree of the numerator is less than the degree of the denominator.

The process involves expressing a complex fraction as a sum of simpler fractions. Each simpler fraction has a denominator that is a factor of the original denominator. For example, in the original exercise, the integrand was \(\frac{1}{x^{2}+6x+8}\). After factoring, it became \((x+2)(x+4)\), leading us to express the integrand as a sum:

• \(\frac{A}{x+2}\)
• \(\frac{B}{x+4}\)

Once decomposed into partial fractions, the individual terms can be integrated separately, making the integration process much simpler.

In our example, using partial fraction decomposition allowed us to handle the integral by substituting \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\) into the equation, thus splitting the integral into two manageable terms.

Linear Equations

In the context of partial fraction decomposition, solving linear equations is crucial for finding the constants in each fraction. Linear equations are equations of the first degree, meaning they involve expressions where the highest power of the variable is one.

When comparing coefficients after expanding the decomposed fractions' equation, you create a system of linear equations. In our exercise, these came from matching the coefficients in the expression \(1 = A(x+4) + B(x+2)\), which expanded to

• \((A+B)x + (4A + 2B)\)

By equating like terms, we obtain:

• \(A + B = 0\)
• \(4A + 2B = 1\)

By solving this system, you determine the values of \(A\) and \(B\). The trick often involves expressing one variable in terms of another, such as \(B = -A\) in this case, then substituting back to find the precise values.

Integration by Substitution

Integration by substitution is a method used to make integration easier by changing the variable of integration to simplify the integral. However, in our particular problem, another straightforward approach was used instead, which tackles the integral directly once we decomposed into partial fractions.

After applying partial fraction decomposition, the integral \(\int \frac{1}{2} \left( \frac{1}{x+2} - \frac{1}{x+4} \right) dx\) used basic integration techniques because the integral of \(\frac{1}{x}\) is readily known. This is an example of how, instead of using substitution, breaking down a complex rational function can lead to straightforward anti-derivatives.

In another context, integration by substitution would be helpful when you have a more complex composition, requiring a change of variable. You'd typically set \(u\) to part of the integrand, substitute the terms, and then integrate concerning \(u\). Upon finding the antiderivative, you revert to the original variable \(x\). Substitution simplifies the integration, especially when the integrand is a product of functions.