Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, evaluate the given integral. $$ \int_{1}^{2} \frac{1}{x^{2}+6 x+8} d x $$

Step-by-Step Solution

Answer

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

Step 1: Factor the Quadratic Expression

First, we need to factor the quadratic expression in the denominator: \(x^2 + 6x + 8\). To factor, look for two numbers that multiply to 8 (the constant term) and add up to 6 (the coefficient of the linear term). These numbers are 2 and 4.Thus, we can write:\[x^2 + 6x + 8 = (x+2)(x+4)\]

Step 2: Split the Fraction Using Partial Fraction Decomposition

Next, we'll express\[\frac{1}{(x+2)(x+4)}\]as a sum of partial fractions. We assume:\[\frac{1}{(x+2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4}\]Multiply through by the denominator \((x+2)(x+4)\) to get:\[1 = A(x+4) + B(x+2)\]Expanding gives:\[1 = Ax + 4A + Bx + 2B\]Combining like terms, we have:\[1 = (A + B)x + (4A + 2B)\]

Step 3: Solve the System of Equations

To find \(A\) and \(B\), compare coefficients:- For the \(x\) terms: \(A + B = 0\)- For the constant terms: \(4A + 2B = 1\)From \(A + B = 0\), we get:\(B = -A\).Substitute \(B = -A\) into the second equation:- \(4A + 2(-A) = 1\)- \(4A - 2A = 1\)- \(2A = 1\)- \(A = \frac{1}{2}\)With \(A = \frac{1}{2}\), substitute back to find \(B\):- \(B = -\frac{1}{2}\)

Step 4: Rewrite the Integral with Partial Fractions

Substitute the values of \(A\) and \(B\) back into the partial fraction decomposition:\[\frac{1}{(x+2)(x+4)} = \frac{1/2}{x+2} - \frac{1/2}{x+4}\]Now rewrite the integral as:\[\int_{1}^{2} \left( \frac{1/2}{x+2} - \frac{1/2}{x+4} \right) dx\]This can be split into two separate integrals:- \[\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} \, dx - \frac{1}{2} \int_{1}^{2} \frac{1}{x+4} \, dx\]

Step 5: Evaluate the Integrals

Now, calculate each integral separately.**First integral:**\[\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} \, dx\] Using the fact that \(\int \frac{1}{x} \, dx = \ln|x| + C\):\[= \frac{1}{2} \left[ \ln|x+2| \right]_{1}^{2}\]\[= \frac{1}{2} \left( \ln(4) - \ln(3) \right)\]**Second integral:**\[\frac{1}{2} \int_{1}^{2} \frac{1}{x+4} \, dx\] Similarly:\[= \frac{1}{2} \left[ \ln|x+4| \right]_{1}^{2}\]\[= \frac{1}{2} \left( \ln(6) - \ln(5) \right)\]

Step 6: Combine and Simplify

Combine both results to get the final value of the original integral:\[\frac{1}{2} \left( \ln(4) - \ln(3) \right) - \frac{1}{2} \left( \ln(6) - \ln(5) \right)\]This simplifies to:\[\frac{1}{2} \left( \ln\left(\frac{4}{3}\right) - \ln\left(\frac{6}{5}\right) \right)\]Using the property \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), this reduces to:\[\frac{1}{2} \ln\left(\frac{4/3}{6/5}\right) = \frac{1}{2} \ln\left(\frac{20}{18}\right) = \frac{1}{2} \ln\left(\frac{10}{9}\right)\]

Step 7: Final Result

Thus, the value of the integral is:\[\frac{1}{2} \ln\left(\frac{10}{9}\right)\]

Key Concepts

Partial Fraction Decomposition

When you're dealing with integrals involving rational functions, especially those with polynomial expressions in the denominator, partial fraction decomposition becomes incredibly handy. It helps split a complex fraction into simpler, easily integrable parts.
In this exercise, we have the rational expression \(\frac{1}{x^2+6x+8}\). The first step was to factor the denominator to become \((x+2)(x+4)\).
This prepares us to express the complex fraction as a sum of simpler fractions with unknown coefficients, assuming the form:

• \(\frac{1}{(x+2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4}\)

Next, multiply both sides by the original denominator to eliminate fractions, which leads to an equation in terms of \(x\):
\[1 = A(x+4) + B(x+2)\]
Expanding and combining like terms allows identifying equations for \(A\) and \(B\) by comparing coefficients. This process simplifies integration because each term can now be integrated more easily than the original complex fraction.

Definite Integrals

Definite integrals are used to calculate the area under a curve between two points on the x-axis. In this problem, the integral \(\int_{1}^{2} \fs \frac{1}{x^2+6x+8} \, dx\) needs to be evaluated between the limits of 1 and 2.
Once you perform partial fraction decomposition, each resulting simpler fraction \(\frac{A}{x+2}\) and \(\frac{B}{x+4}\) can be integrated separately within these limits.
This structure converts the original integral into the sum of two separate integrals:

• \(\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} \, dx\)
• \(- \frac{1}{2} \int_{1}^{2} \frac{1}{x+4} \, dx\)

To evaluate these definite integrals, you calculate the antiderivative of each fraction, substitute the upper and lower bounds, and then subtract accordingly. Definite integrals thus provide a numeric result, representing the net area between the curve and the x-axis, accounting for parts above and below the axis.

Logarithmic Integration

Logarithmic integration is used when integrating functions of the form \(\frac{1}{x+a}\), where the integral results in a logarithmic function. In this exercise, each decomposed fraction, like \(\frac{1}{2}\int \frac{1}{x+2} \, dx\), is a perfect candidate for logarithmic integration.
The general formula for logarithmic integration is

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

For definite integrals, instead of adding an arbitrary constant, apply the limits of integration:
\[\left[ \ln|x+a| \right]_{b}^{c} = \ln|c+a| - \ln|b+a|\]
In our solved problem, after applying the partial fraction decomposition, each integral ultimately took on the form of \(\int \frac{1}{x+a} \, dx\). This application resulted in expressions like \(\frac{1}{2}(\ln(4) - \ln(3))\), which were combined using logarithmic properties to finally solve the definite integral.