A-Level Techniques of integration: worked solution

Question

Find the integral of 9x/((2x - 1)(x + 1)^2) with respect to x.

Step 1: choose the correct partial-fraction form. The denominator has a linear factor (2x − 1) and a repeated linear factor (x + 1)². A repeated factor needs one term for each power, so write 9x/((2x − 1)(x + 1)²) = A/(2x − 1) + B/(x + 1) + C/(x + 1)². Multiply both sides by (2x − 1)(x + 1)²: 9x = A(x + 1)² + B(2x − 1)(x + 1) + C(2x − 1). (*) Step 2: find the three constants. Put x = −1 in (*), which kills the A and B terms: 9(−1) = A(0) + B(0) + C(2(−1) − 1) −9 = −3C, so C = 3. Put x = 1/2 in (*), which kills the B and C terms: 9(1/2) = A(1/2 + 1)² 9/2 = A(3/2)² = (9/4)A, so A = 2. Compare the coefficients of x² on both sides of (*): on the left there is no x², and on the right A(x²) + B(2x²) gives A + 2B, so 0 = A + 2B, hence B = −A/2 = −1. Therefore 9x/((2x − 1)(x + 1)²) = 2/(2x − 1) − 1/(x + 1) + 3/(x + 1)². Step 3: integrate each piece. ∫ 2/(2x − 1) dx = 2 × (1/2)ln|2x − 1| = ln|2x − 1| (the factor 1/2 comes from the derivative of 2x − 1 being 2). ∫ −1/(x + 1) dx = −ln|x + 1|. ∫ 3/(x + 1)² dx = 3∫ (x + 1)^(−2) dx = 3 × (x + 1)^(−1)/(−1) = −3/(x + 1). Step 4: add the three results and include the constant of integration. ∫ 9x/((2x − 1)(x + 1)²) dx = ln|2x − 1| − ln|x + 1| − 3/(x + 1) + c.