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Magazine Chapter 7: Problem 14
Evaluate the integral. \( \displaystyle \int \frac{1}{(x + a)(x + b)}\ dx \)
Short Answer
Expert verified
The integral is \( \frac{1}{b-a} \ln\left|\frac{x+a}{x+b}\right| + C \).
Step by step solution
01
Partial Fraction Decomposition
To evaluate the integral \( \int \frac{1}{(x + a)(x + b)}\ dx \), we first perform partial fraction decomposition. We want to express the integrand \( \frac{1}{(x + a)(x + b)} \) as a sum of partial fractions: \[ \frac{1}{(x + a)(x + b)} = \frac{A}{x + a} + \frac{B}{x + b} \] where \( A \) and \( B \) are constants to be determined.
02
Solve for Constants A and B
Multiply both sides by \((x + a)(x + b)\) to eliminate the denominators: \[ 1 = A(x + b) + B(x + a) \] Expand the right-hand side: \[ 1 = Ax + Ab + Bx + Ba \]Combine like terms: \[ 1 = (A + B)x + (Ab + Ba) \]This gives us the system of equations: \( A + B = 0 \) and \( Ab + Ba = 1 \). Solving \( A = -B \) and substituting into \( Ab + Ba = 1 \) gives us \( Ab - Ab = 1 \), simplifying to \( A = \frac{1}{b-a} \) and \( B = \frac{-1}{b-a} \).
03
Integration of Partial Fractions
Substitute \( A = \frac{1}{b-a} \) and \( B = \frac{-1}{b-a} \) into the partial fractions: \[ \frac{1}{(b-a)} \cdot \left(\frac{1}{x+a} - \frac{1}{x+b} \right) \]The integral becomes: \[ \int \frac{1}{b-a} \left( \frac{1}{x+a} - \frac{1}{x+b} \right) dx \]Separate the integrals: \[ \frac{1}{b-a} \left( \int \frac{1}{x+a} \ dx - \int \frac{1}{x+b} \ dx \right) \]
04
Evaluate Each Integral
Evaluate each integral separately:\( \int \frac{1}{x+a} \ dx = \ln|x+a| + C_1 \)\( \int \frac{1}{x+b} \ dx = \ln|x+b| + C_2 \)Thus, the combined solution is: \[ \frac{1}{b-a} \left( \ln|x+a| - \ln|x+b| \right) + C \] where \( C \) is the constant of integration.
05
Simplify the Logarithmic Expression
Using the logarithm property \( \ln|x+a| - \ln|x+b| = \ln\frac{|x+a|}{|x+b|} \), the solution simplifies to: \[ \frac{1}{b-a} \ln\left|\frac{x+a}{x+b}\right| + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Techniques
Solving integrals can often seem complex, but using various integration techniques simplifies the process. One such approach is **Partial Fraction Decomposition**, especially handy when dealing with rational functions. This method takes an integrand that is a fraction and breaks it into simpler "partial fractions" that you can integrate more comfortably.
This process involves expressing the original function as a sum of simpler fractions. For example, for the integral \(\int \frac{1}{(x + a)(x + b)}\ dx\), we rewrite the single fraction \(\frac{1}{(x + a)(x + b)}\) as the sum of two fractions: \(\frac{A}{x + a} + \frac{B}{x + b}\), where \(A\) and \(B\) are constants.
This conversion makes the function easier to work with and allows us to apply straightforward integration techniques on each separate fraction.
This process involves expressing the original function as a sum of simpler fractions. For example, for the integral \(\int \frac{1}{(x + a)(x + b)}\ dx\), we rewrite the single fraction \(\frac{1}{(x + a)(x + b)}\) as the sum of two fractions: \(\frac{A}{x + a} + \frac{B}{x + b}\), where \(A\) and \(B\) are constants.
This conversion makes the function easier to work with and allows us to apply straightforward integration techniques on each separate fraction.
Logarithmic Integration
Integration often involves finding the antiderivatives of functions. **Logarithmic Integration** is particularly useful when the function to integrate involves fractions where the numerator is the derivative of the denominator, or a close relative of it.
In our problem \(\int \frac{1}{x+a}\ dx\) or \(\int \frac{1}{x+b}\ dx\), we're dealing with basic terms that lead directly to a logarithmic form when integrated. You end up with expressions like \(\ln|x+a| + C_1\) and \(\ln|x+b| + C_2\).
Utilizing properties of logarithms simplifies the solution as well. The expression \(\ln|x+a| - \ln|x+b|\) becomes \(\ln\frac{|x+a|}{|x+b|}\). Thus, the integral of \(\frac{1}{x+a}\) and \(\frac{1}{x+b}\) translates into expressions involving natural logarithms, capturing the continuous growth inherent in these terms.
In our problem \(\int \frac{1}{x+a}\ dx\) or \(\int \frac{1}{x+b}\ dx\), we're dealing with basic terms that lead directly to a logarithmic form when integrated. You end up with expressions like \(\ln|x+a| + C_1\) and \(\ln|x+b| + C_2\).
Utilizing properties of logarithms simplifies the solution as well. The expression \(\ln|x+a| - \ln|x+b|\) becomes \(\ln\frac{|x+a|}{|x+b|}\). Thus, the integral of \(\frac{1}{x+a}\) and \(\frac{1}{x+b}\) translates into expressions involving natural logarithms, capturing the continuous growth inherent in these terms.
Constant of Integration
Whenever you integrate a function, there's the crucial aspect of including the **Constant of Integration**. This constant, denoted typically as \(C\), represents an arbitrary constant added to the indefinite integral.
The necessity for this constant arises because differentiation of a constant is zero, meaning any constant can fit into the original function. Therefore, to capture all potential antiderivatives, this constant is included.
In our case, the inclusion of \(C\) in \(\frac{1}{b-a}\ln\left|\frac{x+a}{x+b}\right| + C\) signifies that there are an infinite number of antiderivative functions, each differing by a constant.
The necessity for this constant arises because differentiation of a constant is zero, meaning any constant can fit into the original function. Therefore, to capture all potential antiderivatives, this constant is included.
In our case, the inclusion of \(C\) in \(\frac{1}{b-a}\ln\left|\frac{x+a}{x+b}\right| + C\) signifies that there are an infinite number of antiderivative functions, each differing by a constant.
