(https://qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2013/Exam-materials/6666_01_que_20160624.pdf) Question 6.(i)

First split up the fraction into partial fractions as this is more easily integrated. It can be seen that the denominator has factors y and (3y+2), therefore we can say the original fraction is equal to partial fractions (A/y)+(B/(3y+2)). The partial fractions are then recombined by multiplying the numerator and denominator of the first fraction by (3y+2), and y for the second fraction. This gives ((3y+2)A+By)/(y(3y+2)), this is of course equal to the original fraction. It can be seen that the original fraction and our recombined partial fraction have equal denominators, therefore we can write 3y-4=(3y+2)A+By. Our next step is to calculate A and B so that we can rewrite the original fraction as a partial fraction. If y is set to 0 the equation 3y-4=(3y+2)A+By becomes -2=A, and if y is set to-2/3 the equation becomes 9=B, we now have both our values for A and B. The integral can now be rewritten as -(2/y)+9/(3y+2), remembering the standard integral int(a/y)=(a)ln(y) we can then integrate our function, therefore the answer is -2ln(y)+3ln(3y+2)+c .