Let f(x) = (x²+4ax+6a²) / ((x+2a)(x+3a)), where a is a positive constant.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] (a) State or imply the form A + B/(x+2a) + C/(x+3a).
2. [1] (a) Use a correct method for finding a coefficient.
3. [1] (a) Obtain one of A=1, B=2a and C =-3a.
4. [1] (a) Obtain a second value.
5. [1] (a) Obtain the third value.
6. [2] (b) Integrate and obtain terms Ax + Bln(x+2a)+Cln(x+3a). B1 for any two terms correct, B2 for all three terms correct. The FT is on A, B and C: x + 2aln(x+2a)-3aln(x+3a).
7. [1] (b) Substitute limits correctly in an integral containing at least 2 terms from the form rx + sln(x+2a)+tln(x+3a). E.g. (a + 2aln(3a) - 3aln(4a)) - (-a + 2aln(a) - 3aln(2a)).
8. [1] (b) Obtain a(2+ln(9/8)) from correct working. Accept equivalent fractions with integers. A0 if a(2+ln(9/8)) is following an error in (a).