Step 1: Understanding the Concept:
The given integral represents a rational algebraic function where the degree of the numerator is less than the degree of the denominator.
Such integrals can be easily evaluated by splitting the integrand into simpler fractions using the method of partial fractions.
Step 2: Key Formula or Approach:
Factorize the quadratic denominator, then express the fraction in the form:
\[ \frac{px + q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \]
Once decomposed, use the standard logarithmic integration rule:
\[ \int \frac{1}{x-a} \, dx = \log |x-a| + C \]
Step 3: Detailed Explanation:
Let the integral be \( I \):
\[ I = \int \frac{2x}{x^2-5x+4} \, dx \]
First, we factorize the quadratic expression in the denominator:
\[ x^2 - 5x + 4 = x^2 - 4x - x + 4 = x(x-4) - 1(x-4) = (x-4)(x-1) \]
So, the integrand becomes:
\[ \frac{2x}{(x-4)(x-1)} \]
Now, we set up the partial fractions decomposition:
\[ \frac{2x}{(x-4)(x-1)} = \frac{A}{x-4} + \frac{B}{x-1} \]
Multiplying both sides by the common denominator \( (x-4)(x-1) \), we get:
\[ 2x = A(x-1) + B(x-4) \]
To find the value of \( A \), we can substitute \( x = 4 \) to eliminate \( B \):
\[ 2(4) = A(4-1) + B(4-4) \]
\[ 8 = 3A \implies A = \frac{8}{3} \]
To find the value of \( B \), we substitute \( x = 1 \) to eliminate \( A \):
\[ 2(1) = A(1-1) + B(1-4) \]
\[ 2 = -3B \implies B = -\frac{2}{3} \]
Substituting the values of \( A \) and \( B \) back into the partial fraction form:
\[ \frac{2x}{(x-4)(x-1)} = \frac{8/3}{x-4} - \frac{2/3}{x-1} \]
Now, substitute this expanded form back into the integral \( I \):
\[ I = \int \left( \frac{8/3}{x-4} - \frac{2/3}{x-1} \right) \, dx \]
Applying the linearity of integration:
\[ I = \frac{8}{3} \int \frac{1}{x-4} \, dx - \frac{2}{3} \int \frac{1}{x-1} \, dx \]
Evaluating the simple logarithmic integrals:
\[ I = \frac{8}{3} \log |x - 4| - \frac{2}{3} \log |x - 1| + C \]
where \( C \) is the constant of integration.
Step 4: Final Answer:
The correct option is (A).