Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Use partial fractions to find the indefinite integral. $$ \int \frac{1}{x^{2}-1} d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(\frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C\).

Step 1: Express the integrand as the sum of partial fractions

The given integrand is a rational function which can be expressed as the sum of partial fractions. In this case, \(x^2-1\) can be factored into \((x-1)(x+1)\), then the fraction can be decomposed to \(A/(x-1) + B/(x+1)\). By equating this to \(\frac{1}{x^2-1}\), coefficients A and B can be solved.

Step 2: Solve for coefficients A and B

Upon equating, we get \(1 = A(x+1) + B(x-1)\). The coefficients A and B can be solved by letting x=-1 and x=1, and solving for A and B respectively. This gives us: A=1/2 and B=-1/2.

Step 3: Integrate the partial fractions

The integral now becomes two separate integrals: \(\int \frac{1/2}{x-1} dx - \int \frac{1/2}{x+1} dx\), where each fraction can be integrated directly using the formula for the integral of 1/x, which is ln|x|.

Step 4: Write down the final solution

The integration of the two separate fractions gives \(\frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C\), where C stands for the constant of integration.