Calculate the integral of (3x+3)/(2x^2+3x) between the limits 39 and 3

The first step is to split the fraction into 2 separate fractions using partial fractions techniques. Write 3x+3/2x^2+3x as A/x + B/2x+3 and solve to get A = 1, B = 1. We have now converted 3x+3/2x^2+3x into two much simpler fractions, 1/x + 1/2x+3

The next step is to integrate and we can recognise that both fractions are of the form, diferential of the denominator (or fraction of it), over the denominator. This means that by recognition we can integrate to get 0.5ln(2x+3) + lnx. Substituting values in and using natural logarithm laws we can come to the conclusion that the answer is ln39