Let f(x) = (2x² + 17x - 17) / ((1 + 2x)(2-x)²).

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. B1 State or imply the form A/(1+2x) + B/(2-x) + C/(2-x)² [Alternative form: A/(1+2x) + (Dx+E)/(2-x)²]
2. M1 Use a correct method for finding a coefficient [e.g. A(2-x)² + B(1 + 2x)(2 – x) + C(1 + 2x) = 2x² + 17x-17 and compare coefficients or substitute for x. A(2 – x)³ + B(1 + 2x)(2 – x)² + C(1 + 2x)(2 - x) = 2x² + 17x – 17 scores M0.]
3. A1 Obtain one of A = -4, B = -3 and C = 5
4. A1 Obtain a second value
5. A1 Obtain the third value [Extra term in partial fractions, then B0 unless recover at end. Allow the marks for any constants found correctly. Missing terms in partial fractions, B0 but M1A1 is available for a correct method that obtains at least one correct constant (e.g. cover-up rule) Max 2/5. Ignore any substitution back into their original expression. If alternative form used: A = -4, D = 3 and E = -1.]
6. B1FT Integrate and obtain terms -2ln(1+2x) + 3ln(2-x) + 5/(2-x) [The FT is on correct use of their A, B and C; or on A, D and E. If using the A, D, E form then B1 for the A term, but no further marks until partial fractions are used to split the second term or they use integration by parts to obtain ∫(Dx+E)/(2-x)² dx for the 2nd B1 and 3rd B1 for correct completion. B0FT, B0FT, B0FT if they place their A, B, C with incorrect denominators.]
7. M1 Substitute limits correctly in an integral with two terms (obtained correctly) of the form aln(1+2x) + bln(2-x) + c/(2-x), where abc≠0 [Condone minor slips in substitution. Exact substitution required.]
8. A1 Obtain answer (5/2) - ln 72 after full and correct working [AG – evidence of some correct work to combine or simplify logs is required e.g. allow from -ln 9 + ln(1/8) or -ln 2³ - ln 3².]