Question

Find : \( \int \frac{3x - 1}{\sqrt{x^2 - 4x}} dx \)

Hint:

Splitting the numerator allows you to solve the radical part using simple substitution.
Completing the square is the standard way to handle the remaining constant numerator part.

Answer 1

Concept:
• Integrals of the form \( \int \frac{px + q}{\sqrt{ax^2 + bx + c}} dx \).
• Substitute \( px + q = A \frac{d}{dx}(ax^2 + bx + c) + B \).
• Standard Integral: \( \int \frac{1}{\sqrt{x^2 - a^2}} dx = \log |x + \sqrt{x^2 - a^2}| + C \).

Step 1: Express numerator in terms of the derivative of the quadratic
Let \( 3x - 1 = A(2x - 4) + B \).
Comparing coefficients of \( x \): \( 2A = 3 \implies A = 3/2 \).
Comparing constants: \( -4A + B = -1 \implies -4(3/2) + B = -1 \implies -6 + B = -1 \implies B = 5 \).

Step 2: Split the integral
\[ I = \frac{3}{2} \int \frac{2x - 4}{\sqrt{x^2 - 4x}} dx + 5 \int \frac{1}{\sqrt{x^2 - 4x}} dx \]
Let \( I = I_1 + I_2 \).

Step 3: Evaluate the first integral \( I_1 \)
For \( I_1 \), put \( x^2 - 4x = t \), then \( (2x - 4)dx = dt \).
\[ I_1 = \frac{3}{2} \int t^{-1/2} dt = \frac{3}{2} [2\sqrt{t}] = 3\sqrt{x^2 - 4x} \]

Step 4: Evaluate the second integral \( I_2 \)
Complete the square: \( x^2 - 4x = (x - 2)^2 - 4 \).
\[ I_2 = 5 \int \frac{dx}{\sqrt{(x-2)^2 - 2^2}} = 5 \log |(x - 2) + \sqrt{x^2 - 4x}| \]

Step 5: Combine the results
\[ I = 3\sqrt{x^2 - 4x} + 5 \log |x - 2 + \sqrt{x^2 - 4x}| + C \]