Problem 16 - College Algebra and Calculus: An Applied Approach

Question

Use partial fractions to find the indefinite integral. $$ \int \frac{-4}{x^{2}-4} d x $$

Step-by-Step Solution

Answer

\[-\ln |x-2| + \ln |x + 2 | + C\]

Step 1: Decomposition into Partial Fractions

Rewrite the rational function as the sum of its simpler fractions. To do this, it's necessary to factorize the denominator. \(x^{2}-4\) can be factorized into \((x-2)(x+2)\). Hence, the integrand can be rewritten using partial fraction decomposition as: \[-\frac{4}{x^{2}-4} = \frac{A}{x-2} + \frac{B}{x+2}\]. Multiplying through by the common denominator gives \(-4 = A(x+2) + B(x-2)\]. Let's solve for A and B by comparing coefficients.

Step 2: Solve for A and B

Setting \(x=2\) in the equation \(-4 = A(x+2) + B(x-2)\) gives \(A(4) = -4\), hence \(A=-1\). Setting \(x=-2\) gives \(B(-4) = -4\), hence \(B=1\). Hence now, the integral becomes: \[ \int \frac{-1}{x-2} + \frac{1}{x+2} dx \].

Step 3: Integrate the Simplified Fraction

Now, the integration can be performed on each simpler fraction separately: \[= - \int \frac{1}{x-2} dx + \int \frac{1}{x+2} dx\] which are in the form of \(\int \frac{1}{x-a} dx\), whose integral is \(\ln |x-a|\), where \(\ln\) is the natural logarithm. Therefore, the final answer is: \[-\ln |x-2| + \ln |x + 2 | + C\]