Problem 27 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

\(-\frac{1}{2}\ln|x|+\frac{1}{2}\ln|x-4| + C\)

Step 1: Rewrite the denominator

First, factorize the denominator, which is a quadratic function, into two linear functions. In this case, \(x^{2} - 4x\) can be written as \(x(x - 4)\).

Step 2: Decomposition into Partial Fractions

\(\frac{x+2}{x(x-4)}\) can be split into two simpler fractions, let's say \(\frac{A}{x} + \frac{B}{x-4}\). To find A and B, clear the fraction by multiplying both sides of the equation by the common denominator x(x-4). This gives: \(x+2=A(x-4)+Bx\). Now to solve for A and B, you select convenient values for x to isolate each variable. For instance, choosing \(x=0\) yields \(A=-\frac{1}{2}\) and choosing \(x=4\) gives \(B=\frac{1}{2}\). Thus, this fraction is decomposed to: \(-\frac{1}{2} \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x-4}\).

Step 3: Evaluate the Integral

Now that the expression is simplified, find the indefinite integral: \(\int [-\frac{1}{2} \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x-4}] dx\). The integral of a sum is the sum of the integrals, so this can be written as: \(-\frac{1}{2}\int \frac{1}{x} dx + \frac{1}{2}\int \frac{1}{x-4} dx\). This results in: \(-\frac{1}{2} \ln |x| +\frac{1}{2} \ln |x-4| + C\), where C is the constant of integration.