Problem 2 - Active Calculus

Question

Calculate the integral: \(\int \frac{1}{(x+6)(x+8)} d x=\) ___________________

Step-by-Step Solution

Answer

\(\frac{1}{2} \ln \left| \frac{x+6}{x+8} \right| + C\).

Step 1 - Partial Fraction Decomposition

Rewrite the integrand \(\frac{1}{(x+6)(x+8)}\) by using partial fractions. Assume \(\frac{1}{(x+6)(x+8)} = \frac{A}{x+6} + \frac{B}{x+8}\).

Step 2 - Determine Constants \(A\) and \(B\)

Multiply through by \((x+6)(x+8)\) to get \(1 = A(x+8) + B(x+6)\). Solve for \(A\) and \(B\) by substituting convenient values for \(x\).

Step 3 - Solve for \(A\)

Set \(x = -8\) to find \(A\): \(1 = A(-8+8) + B(-8+6)\) simplifies to \(1 = B(-2)\), so \(B = -\frac{1}{2}\).

Step 4 - Solve for \(B\)

Set \(x = -6\) to find \(B\): \(1 = A(-6+8) + B(-6+6)\) simplifies to \(1 = 2A\), so \(A = \frac{1}{2}\).

Step 5 - Rewrite the Integrand

Substitute \(A\) and \(B\) into the partial fractions: \(\frac{1}{(x+6)(x+8)} = \frac{1/2}{x+6} - \frac{1/2}{x+8}\).

Step 6 - Integrate

Integrate each term separately: \(\int \frac{1/2}{x+6} dx - \int \frac{1/2}{x+8} dx = \frac{1}{2} \ln |x+6| - \frac{1}{2} \ln |x+8| + C\).

Step 7 - Combine Logarithms

Combine the logarithmic terms: \(\frac{1}{2} (\ln |x+6| - \ln |x+8|) + C = \frac{1}{2} \ln \left| \frac{x+6}{x+8} \right| + C\).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a crucial technique in integral calculus. It allows us to break down a complex rational function into simpler fractions that are easier to integrate.
Let's start by rewriting the given integrand \(\frac{1}{(x+6)(x+8)}\) using partial fractions.
First, assume that: \(\frac{1}{(x+6)(x+8)} = \frac{A}{x+6} + \frac{B}{x+8}\). Here, \A\ and \B\ are constants that we need to determine.
Multiplying through by the denominator \( (x+6)(x+8) \), we get: \1 = A(x+8) + B(x+6)\.
To find the values of \A\ and \B\, we can choose values for \x\ that simplify our equation. Setting \x = -8\ makes the term \A(x+8)\ zero, leaving us with: \1 = B(-8+6) \Rightarrow 1 = -2B \Rightarrow B = -\frac{1}{2}\.
Similarly, setting \x = -6\ cancels out the \B(x+6)\ term: \1 = A(-6+8) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}\.
Now we rewrite our integrand using these values: \frac{1}{(x+6)(x+8)} = \frac{1/2}{x+6} - \frac{1/2}{x+8}\.

Integration Techniques

Integration techniques involve various methods to solve integrals, and partial fraction decomposition is one such method.
Once we have simplified our integrand, we can integrate each part separately. Here, we need to integrate \(\frac{1/2}{x+6}\) and \(-\frac{1/2}{x+8}\).
The integral of \(\frac{1}{x+k}\) is \(\text{ln} |x+k| + C\), where \C\ is the constant of integration.
Therefore, integrating each term separately, we get: \ ∫ \frac{1/2}{x+6} dx - ∫ \frac{1/2}{x+8} dx\. The first integral results in \(\frac{1}{2} \text{ln} | x+6 |\) and the second integral results in \(\frac{1}{2} \text{ln} | x+8 |\).
So, the result of our integral becomes: \(\frac{1}{2} \text{ln} |x+6| - \frac{1}{2} \text{ln} |x+8| + C\).
Notice how breaking down the complex fraction into simpler parts allowed us to use basic integration techniques on each simpler fraction.

Logarithmic Integration

Logarithmic integration is useful when dealing with integrals of the form \(\frac{1}{x+k}\).
In our example, after partial fraction decomposition, we had two simple logarithmic integrals.
Combining logarithms is another handy trick in integral calculus. Our solution, \(\frac{1}{2} \text{ln} | x+6 | - \frac{1}{2} \text{ln} | x+8 | + C\), can be simplified.
We use the logarithmic property: \ \text{ln}a - \text{ln}b = \text{ln}(\frac{a}{b}) \.
Applying this property, we group the logarithmic terms: \(\frac{1}{2} (\text{ln} |x+6| - \text{ln} |x+8|) + C \). This simplifies to: \(\frac{1}{2} \text{ln} \frac{|x+6|}{|x+8|} + C \).
Thus, the final answer for our integral is: \(\frac{1}{2} \text{ln} \frac{|x+6|}{|x+8|} + C \).
This example shows how logarithmic integration, combined with properties of logarithms, simplifies the integration process.