A-Level Techniques of integration: worked solution

Question

Evaluate the integral of (x^2 - 3x - 1)/(x^2 - 4x + 1) with respect to x.

Step 1 - notice the top and the bottom have the same degree, so divide first. When the numerator is not of lower degree than the denominator, you must divide before trying any other technique. Compare x² - 3x - 1 with x² - 4x + 1. Subtract the denominator from the numerator to see what is left over: (x² - 3x - 1) - (x² - 4x + 1) = -3x + 4x - 1 - 1 = x - 2. So x² - 3x - 1 = 1 × (x² - 4x + 1) + (x - 2), which means (x² - 3x - 1)/(x² - 4x + 1) = 1 + (x - 2)/(x² - 4x + 1). Step 2 - look at the leftover fraction and test it for the f'(x)/f(x) pattern. Differentiate the denominator: d/dx (x² - 4x + 1) = 2x - 4 = 2(x - 2). So the numerator x - 2 is exactly half the derivative of the denominator: x - 2 = (1/2)(2x - 4). Therefore (x - 2)/(x² - 4x + 1) = (1/2) × (2x - 4)/(x² - 4x + 1), which is (1/2) × f'(x)/f(x) and integrates to (1/2) ln|f(x)|. Step 3 - integrate the two pieces. ∫ (x² - 3x - 1)/(x² - 4x + 1) dx = ∫ 1 dx + (1/2) ∫ (2x - 4)/(x² - 4x + 1) dx = x + (1/2) ln|x² - 4x + 1| + C. Note on the modulus: x² - 4x + 1 = 0 at x = 2 ± √3, so the quadratic is negative between those two values. The modulus signs are needed, not optional.