Step 1: Understanding the Question:
The given problem requires evaluating an integral of a rational function where the degree of the numerator is less than the degree of the denominator.
The denominator is a quadratic expression that can be factored, making it a classic case for partial fraction decomposition.
Step 2: Key Formula or Approach:
1. Factor the denominator into linear factors: \(x^2 - 5x + 4 = (x-a)(x-b)\).
2. Decompose the fraction using the form: \( \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \).
3. Use the standard integral formula: \( \int \frac{1}{x-k} dx = \ln|x-k| + C \).
Step 3: Detailed Explanation:
First, factor the denominator:
\[ x^2 - 5x + 4 = (x-1)(x-4) \]
Set up the partial fraction decomposition:
\[ \frac{2x}{(x-1)(x-4)} = \frac{A}{x-1} + \frac{B}{x-4} \]
Multiply through by the common denominator:
\[ 2x = A(x-4) + B(x-1) \]
Solve for coefficients by substituting suitable values of \(x\):
If \(x = 4\):
\[ 2(4) = B(4-1) \Rightarrow 8 = 3B \Rightarrow B = \frac{8}{3} \]
If \(x = 1\):
\[ 2(1) = A(1-4) \Rightarrow 2 = -3A \Rightarrow A = -\frac{2}{3} \]
Now, substitute the values back into the integral:
\[ \int \left( \frac{-2/3}{x-1} + \frac{8/3}{x-4} \right) dx \]
Perform the integration term by term:
\[ = -\frac{2}{3}\ln|x-1| + \frac{8}{3}\ln|x-4| + C \]
Rearrange the terms for the final format:
\[ = \frac{8}{3}\ln|x-4| - \frac{2}{3}\ln|x-1| + C \]
Step 4: Final Answer:
The integral evaluates to \( \frac{8}{3}\ln|x-4| - \frac{2}{3}\ln|x-1| + C \).