Given that y > 0, find ∫((3y - 4)/y(3y + 2)) dy (taken from the Edexcel C4 2016 paper)

This can't be integrated directly as y appears in the numerator and denominator. This is an indication that you must integrate by parts. A/y + B/(3y+2) = (3y - 4)/y(3y + 2) A and B must be found. Multiply by the denominators to get A and B on their own. All denominators should disappear. A(3y + 2) + B(y) = 3y - 4 To find A and B, find the value of y that would make the value of the bracket 0. This will cancel out A or B. Here, it's y = 0 and y = -2/3. Plug each value into the equation. y = 0                   y = -2 / 3 2A = -4                 -2B / 3 = -6 A = -2                  B = 9 Now that A and B have been found, the integral looks like ∫(9 / 3y + 2) - (2 / y) dy. The equation can now be integrated, giving the answer 27ln(3y + 2) - 2ln(y) (+c)