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To solve the integral [tex]\(\int \frac{1}{x^2 - a^2} \, dx\)[/tex], we start by noticing that the integrand has the form of a rational function where the denominator is a difference of squares. Here's a step-by-step way to integrate it:
1. Rewrite the integrand:
Given the integrand [tex]\(\frac{1}{x^2 - a^2}\)[/tex], we recognize that [tex]\(x^2 - a^2\)[/tex] can be factored as a difference of squares:
[tex]\[ x^2 - a^2 = (x - a)(x + a) \][/tex]
Therefore, the integrand can be rewritten as:
[tex]\[ \frac{1}{(x - a)(x + a)} \][/tex]
2. Partial fraction decomposition:
We can decompose [tex]\(\frac{1}{(x - a)(x + a)}\)[/tex] into partial fractions. We write:
[tex]\[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a} \][/tex]
To find the constants [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we set up the equation:
[tex]\[ 1 = A(x + a) + B(x - a) \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can substitute convenient values for [tex]\(x\)[/tex]:
- For [tex]\(x = a\)[/tex], the equation becomes [tex]\(1 = A(a + a) + B(a - a)\)[/tex], which simplifies to [tex]\(1 = 2aA\)[/tex]. Thus:
[tex]\[ A = \frac{1}{2a} \][/tex]
- For [tex]\(x = -a\)[/tex], the equation becomes [tex]\(1 = A(-a + a) + B(-a - a)\)[/tex], which simplifies to [tex]\(1 = -2aB\)[/tex]. Thus:
[tex]\[ B = \frac{-1}{2a} \][/tex]
3. Rewrite the integrand using partial fractions:
Substitute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the partial fraction decomposition:
[tex]\[ \frac{1}{(x - a)(x + a)} = \frac{1}{2a} \cdot \frac{1}{x - a} - \frac{1}{2a} \cdot \frac{1}{x + a} \][/tex]
4. Integrate each term separately:
The integral splits into two simpler integrals:
[tex]\[ \int \frac{1}{(x - a)(x + a)} \, dx = \int \left( \frac{1}{2a} \frac{1}{x - a} - \frac{1}{2a} \frac{1}{x + a} \right) \, dx \][/tex]
This can be integrated term-by-term:
[tex]\[ \int \frac{1}{2a} \frac{1}{x - a} \, dx - \int \frac{1}{2a} \frac{1}{x + a} \, dx \][/tex]
5. Evaluate the integrals:
- For [tex]\(\int \frac{1}{x - a} \, dx\)[/tex], the result is [tex]\(\ln |x - a|\)[/tex].
- For [tex]\(\int \frac{1}{x + a} \, dx\)[/tex], the result is [tex]\(\ln |x + a|\)[/tex].
So we have:
[tex]\[ \frac{1}{2a} \ln |x - a| - \frac{1}{2a} \ln |x + a| \][/tex]
6. Combine the logarithms:
Using properties of logarithms, specifically [tex]\(\ln a - \ln b = \ln \frac{a}{b}\)[/tex], we get:
[tex]\[ \frac{1}{2a} \left( \ln |x - a| - \ln |x + a| \right) = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| \][/tex]
Therefore, the integral [tex]\(\int \frac{1}{x^2 - a^2} \, dx\)[/tex] is:
[tex]\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
1. Rewrite the integrand:
Given the integrand [tex]\(\frac{1}{x^2 - a^2}\)[/tex], we recognize that [tex]\(x^2 - a^2\)[/tex] can be factored as a difference of squares:
[tex]\[ x^2 - a^2 = (x - a)(x + a) \][/tex]
Therefore, the integrand can be rewritten as:
[tex]\[ \frac{1}{(x - a)(x + a)} \][/tex]
2. Partial fraction decomposition:
We can decompose [tex]\(\frac{1}{(x - a)(x + a)}\)[/tex] into partial fractions. We write:
[tex]\[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a} \][/tex]
To find the constants [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we set up the equation:
[tex]\[ 1 = A(x + a) + B(x - a) \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can substitute convenient values for [tex]\(x\)[/tex]:
- For [tex]\(x = a\)[/tex], the equation becomes [tex]\(1 = A(a + a) + B(a - a)\)[/tex], which simplifies to [tex]\(1 = 2aA\)[/tex]. Thus:
[tex]\[ A = \frac{1}{2a} \][/tex]
- For [tex]\(x = -a\)[/tex], the equation becomes [tex]\(1 = A(-a + a) + B(-a - a)\)[/tex], which simplifies to [tex]\(1 = -2aB\)[/tex]. Thus:
[tex]\[ B = \frac{-1}{2a} \][/tex]
3. Rewrite the integrand using partial fractions:
Substitute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the partial fraction decomposition:
[tex]\[ \frac{1}{(x - a)(x + a)} = \frac{1}{2a} \cdot \frac{1}{x - a} - \frac{1}{2a} \cdot \frac{1}{x + a} \][/tex]
4. Integrate each term separately:
The integral splits into two simpler integrals:
[tex]\[ \int \frac{1}{(x - a)(x + a)} \, dx = \int \left( \frac{1}{2a} \frac{1}{x - a} - \frac{1}{2a} \frac{1}{x + a} \right) \, dx \][/tex]
This can be integrated term-by-term:
[tex]\[ \int \frac{1}{2a} \frac{1}{x - a} \, dx - \int \frac{1}{2a} \frac{1}{x + a} \, dx \][/tex]
5. Evaluate the integrals:
- For [tex]\(\int \frac{1}{x - a} \, dx\)[/tex], the result is [tex]\(\ln |x - a|\)[/tex].
- For [tex]\(\int \frac{1}{x + a} \, dx\)[/tex], the result is [tex]\(\ln |x + a|\)[/tex].
So we have:
[tex]\[ \frac{1}{2a} \ln |x - a| - \frac{1}{2a} \ln |x + a| \][/tex]
6. Combine the logarithms:
Using properties of logarithms, specifically [tex]\(\ln a - \ln b = \ln \frac{a}{b}\)[/tex], we get:
[tex]\[ \frac{1}{2a} \left( \ln |x - a| - \ln |x + a| \right) = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| \][/tex]
Therefore, the integral [tex]\(\int \frac{1}{x^2 - a^2} \, dx\)[/tex] is:
[tex]\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
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