Problem 63 - Calculus Single Variable

Question

Assume that \(a\) is a positive constant and that \(|x|

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{2a} \ln \left(\frac{a+x}{a-x}\right) + C\).

Step 1: Analyze the Denominator

The denominator of the integrand is \(a^2 - x^2\), which can be factored as \((a-x)(a+x)\). This is the difference of squares, which is a common form for applying partial fraction decomposition.

Step 2: Set Up Partial Fraction Decomposition

Express \(\frac{1}{a^2 - x^2}\) as the sum of two fractions: \(\frac{1}{(a-x)(a+x)} = \frac{A}{a-x} + \frac{B}{a+x}\). We need to determine the constants \(A\) and \(B\).

Step 3: Solve for Constants A and B

Multiply through by the denominator \((a-x)(a+x)\) to clear the fractions: \(1 = A(a+x) + B(a-x)\). Expanding, we get \(A(a+x) + B(a-x) = Aa + Ax + Ba - Bx\). Set this equal to 1 and equate coefficients to find \(A\) and \(B\):\\[-1: Ax - Bx = 0 \rightarrow A = B\] \\[a: Aa + Ba = 1 \rightarrow a(A + B) = 1\].\Substitute \(A = B\) into the second equation: \(a(2A) = 1 \rightarrow A = \frac{1}{2a}\). Therefore, \(B = \frac{1}{2a}\) as well.

Step 4: Rewrite the Integral Using Partial Fractions

Substitute back into the integral: \[\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\int \frac{dx}{a-x} + \frac{1}{2a} \int \frac{dx}{a+x}.\]

Step 5: Solve Each Integral

Each of these integrals is the standard natural logarithm form, so:\\[\int \frac{dx}{a-x} = -\ln |a-x| + C_1,\]\[\int \frac{dx}{a+x} = \ln |a+x| + C_2.\\]Thus, combining results gives: \[\frac{1}{2a}(-\ln |a-x| + \ln |a+x|).\]

Step 6: Simplify the Expression

Combine the logarithms using the property \(\ln a - \ln b = \ln{(\frac{a}{b})}\): \[\frac{1}{2a} \left( \ln |a+x| - \ln |a-x| \right) = \frac{1}{2a} \ln \left(\frac{a+x}{a-x}\right).\] Add \(C\) for the constant of integration.

Key Concepts

Partial Fractions

Partial fractions are a powerful tool used in calculus, especially when dealing with complex rational expressions. The idea is to express a fraction as a sum of simpler fractions, which is often easier to integrate. For example, when you encounter a fraction like \( \frac{1}{a^2 - x^2} \), it can be broken down into simpler parts. This is because the denominator can be factored into \((a-x)(a+x)\), a product of two linear factors.

• The general principle is that any proper rational expression can be decomposed into partial fractions.
• This is done by assuming the expression function is a fraction of the simplest terms, usually expressed as \( \frac{A}{a-x} + \frac{B}{a+x} \) in cases of linear factors.
• The constants \(A\) and \(B\) are found by algebraic manipulation, such as setting up the equation to equal the original fraction and solving for the unknowns.

In our exercise, we see this method applied by setting \( \frac{1}{a^2-x^2}=\frac{A}{a-x} + \frac{B}{a+x} \). Solving for \(A\) and \(B\) helps simplify the integration process.

Integral Calculus

Integral calculus is concerned with determining the integral, or antiderivative, of a function. In essence, it finds the "accumulated value" that results from a function, often related to area under a curve. This can be especially useful in problems requiring the total length, area, volume, or other summation over an interval.

• To integrate using a partial fraction setup involves integrating each simpler fraction separately.
• This process can turn a challenging integral problem into manageable steps.

In the original exercise, after partial fraction decomposition, the integral is represented as two distinct integrals \( \frac{1}{2a} \int \frac{dx}{a-x} + \frac{1}{2a} \int \frac{dx}{a+x} \), which are both much simpler to solve independently as they follow a pattern that leads naturally to logarithmic integration.

Logarithmic Integration

Logarithmic integration appears frequently when dealing with fractions that have linear denominators. This technique involves knowing the integral of simple forms like \( \int \frac{1}{x} \, dx = \ln |x| + C \). Here, the integral results in natural logarithms, which are straightforward to compute.

• The key to logarithmic integration is applying the natural logarithm identity correctly once the correct form is achieved.
• Each fraction results in a natural logarithm; understanding these forms aids in fast recognition and accurate calculation of the integral.

In the exercise, each integral \( \int \frac{dx}{a-x} \) and \( \int \frac{dx}{a+x} \) converts to \( -\ln |a-x| + C_1 \) and \( \ln |a+x| + C_2 \), respectively. This helps to form the combined result \( \frac{1}{2a} \ln \left(\frac{a+x}{a-x}\right) + C \), demonstrating the usefulness of logarithmic techniques.

Constant of Integration

In calculus, whenever you find an indefinite integral, it's crucial to include the constant of integration, commonly denoted as \(C\). This acknowledges that there are infinitely many possible antiderivatives for any given function, differing only by a constant value.

• The constant of integration represents an unknown, constant shift in the antiderivative curve.
• Ignoring this constant could result in a solution that does not satisfy all possible initial conditions.

In our exercise, once the integration process is fully resolved leading to \( \frac{1}{2a} \ln \left(\frac{a+x}{a-x}\right) \), \(C\) is added to denote that any constant could be added to the solution, making it a general solution. This constant is critical when applying the general solution to specific conditions or boundary values in real-world contexts.