Questions on integrating rational functions

Question 1

Find ∫2/(3x + 5) dx.

Accepted Answer

(2/3) ln|3x + 5| + c. M1 for a logarithm of the bracket, A1 for the full answer. Answers of 2 ln|3x + 5| + c omit the division by the inner coefficient 3.

Question 2

Find ∫5/(2x − 1) dx, and ∫3/(x + 2) dx.

Accepted Answer

(5/2) ln|2x − 1| + c and 3 ln|x + 2| + c. M1 for the logarithmic form, A1 for the first integral, B1 for the second.

Question 3

For each of ∫5/(x − 4)³ dx, ∫(2x + 3)/(x² + 3x + 8) dx and ∫(x² + 4)/(x + 1) dx, state which one of these should be done first: reverse the power rule, use the f′/f logarithm, or divide. Give a reason in each case. You are not asked to evaluate the integrals.

Accepted Answer

In the first, the top is a constant and the bottom is a bracket raised to a power, so write it as 5(x − 4)⁻³ and reverse the power rule. No logarithm can appear: the derivative of (x − 4)³ is 3(x − 4)², which the top is not a multiple of. B1. In the second, differentiating x² + 3x + 8 gives exactly 2x + 3, the top, so the f′/f pattern applies and the answer is a logarithm of the bottom. B1. Partial fractions are not on offer there in any case, since the discriminant is 9 − 32 = −23 and the bottom has no real linear factors. In the third, the top has degree 2 and the bottom degree 1, so the fraction is improper and the first move is to divide: x² + 4 = (x + 1)(x − 1) + 5, giving x − 1 + 5/(x + 1). B1.

Question 4

A student writes ∫(x + 5)/((x + 1)(x + 3)) dx = 2 ln|x + 1| + ln|x + 3| + c. By differentiating the student's answer, show that it is wrong, and give the correct integral.

Accepted Answer

Differentiate the student's answer. Each logarithm gives the reciprocal of its bracket, both inner coefficients being 1, so the derivative is 2/(x + 1) + 1/(x + 3). Over the common denominator that is [2(x + 3) + (x + 1)]/((x + 1)(x + 3)) = (3x + 7)/((x + 1)(x + 3)). M1 for differentiating and combining over the common denominator. The integrand's numerator is x + 5, not 3x + 7, so the answer is wrong; at x = 0 the two read 7/3 and 5/3. A1 for the comparison. The split has to satisfy A(x + 3) + B(x + 1) = x + 5, and x = −1 gives 2A = 4 while x = −3 gives −2B = 2, so A = 2 and B = −1 and the integral is 2 ln|x + 1| − ln|x + 3| + c. B1 for the correct answer.

Question 5

Find ∫(5x + 8)/((x + 1)(x + 2)) dx.

Accepted Answer

Split the fraction first. Writing (5x + 8)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2) and substituting the roots x = −1 and x = −2 gives A = 3 and B = 2. Each piece is a logarithm, so the integral is 3 ln|x + 1| + 2 ln|x + 2| + c. M1 for setting up partial fractions, A1 for A = 3 and B = 2, M1 for integrating to logarithms, A1 for the answer.

Question 6

Find ∫x/(x² + 4) dx, and explain why partial fractions cannot be used here.

Accepted Answer

The numerator is half the derivative of the denominator, so the f′/f pattern applies and the integral is ½ ln(x² + 4) + c. No modulus is needed, since x² + 4 is always positive. Partial fractions require the denominator to factorise into real linear brackets, and x² + 4 has discriminant −16, so it has no real roots. M1 for recognising the f′/f pattern, A1 for the integral, B1 for the reason partial fractions fail.

Question 7

Find the exact value of ∫₀¹ 2/((x + 1)(x + 2)) dx, giving your answer as a single logarithm.

Accepted Answer

Split first. 2/((x + 1)(x + 2)) = 2/(x + 1) − 2/(x + 2), so the integral is [2 ln(x + 1) − 2 ln(x + 2)] from 0 to 1. That comes to 2(ln 2 − ln 3) − 2(ln 1 − ln 2) = 4 ln 2 − 2 ln 3 = 2 ln(4/3) = ln(16/9). M1 for splitting into partial fractions, A1 for the two fractions, M1 for integrating to logarithms, A1 for substituting both limits, A1 for ln(16/9). The question asks for an exact value, so leave the logarithms in place rather than writing 0.575, and evaluate both brackets at the lower limit.

Question 8

Find ∫3/(2x − 5)⁴ dx.

Accepted Answer

The bottom is a bracket raised to a power, not a linear bracket, so no logarithm is involved. Rewrite the integrand as a power, 3(2x − 5)⁻⁴, and reverse the power rule: raise the index to −3, divide by the new index −3, and divide again by the inner coefficient 2. That gives 3(2x − 5)⁻³/(−6) = −1/(2(2x − 5)³) + c. M1 for rewriting as a negative power, M1 for raising the index and dividing by it and by the inner coefficient, A1 for the answer. A logarithm appears only where the top is the bottom's derivative up to a constant, and the derivative of (2x − 5)⁴ is 8(2x − 5)³, which is nowhere in sight.

Question 9

Find ∫(5x + 1)/(x − 2)² dx.

Accepted Answer

A repeated bracket takes both powers, so write (5x + 1)/(x − 2)² = A/(x − 2) + B/(x − 2)² and multiply up: 5x + 1 = A(x − 2) + B. Substituting x = 2 gives B = 11, and comparing the coefficients of x gives A = 5. The two pieces then integrate by different rules. 5/(x − 2) is a logarithm, giving 5 ln|x − 2|, while 11/(x − 2)² is the power 11(x − 2)⁻², integrating to −11/(x − 2). So the integral is 5 ln|x − 2| − 11/(x − 2) + c. M1 for the repeated-factor template, A1 for B = 11, A1 for A = 5, M1 for integrating one logarithm and one negative power, A1 for the answer. Watch the sign, since integrating the positive term 11(x − 2)⁻² gives a negative result and then the minus in front of 11/(x − 2) is easy to drop.

Question 10

Find ∫(x² + 7x + 17)/(x + 3) dx.

Accepted Answer

Compare the degrees before choosing anything: the top has degree 2 and the bottom degree 1, so the fraction is improper and has to be divided down first. M1 for dividing rather than reaching for a logarithm. Dividing, x² + 7x + 17 = (x + 3)(x + 4) + 5, so the integrand is x + 4 + 5/(x + 3). A1 for the quotient and the remainder. The three pieces go separately: x integrates to x²/2, the constant 4 to 4x, and the remainder is a constant over a linear bracket whose inner coefficient is 1, giving 5 ln|x + 3|. M1 for integrating all three. So the integral is x²/2 + 4x + 5 ln|x + 3| + c. A1 for the answer.

Question 11

Find the exact value of ∫₀¹ 5/(2x − 5) dx, giving your answer in the form k ln(a/b).

Accepted Answer

A constant over a linear bracket integrates to a logarithm, divided by the inner coefficient: [(5/2) ln|2x − 5|] from 0 to 1. M1 for the logarithmic form with the division by 2. The bracket 2x − 5 runs from −5 at the lower limit to −3 at the upper and is negative throughout, so |2x − 5| = 5 − 2x across the interval and the logarithms taken are of 5 and 3. A1 for the modulus with that reason. Substituting, the value is (5/2)(ln 3 − ln 5). M1 for both limits. The subtraction law compresses it to (5/2) ln(3/5). A1 for the exact value. The answer is negative, as it must be: 5/(2x − 5) is negative across [0, 1], so the region lies below the axis.

Question 12

Find ∫(x² + 3x + 3)/((x + 1)(x + 2)) dx.

Accepted Answer

The fraction is improper, since numerator and denominator both have degree 2, so divide before splitting. The denominator expands to x² + 3x + 2, and the numerator is that plus 1, so the integrand is 1 + 1/((x + 1)(x + 2)). Partial fractions on the remainder give 1/(x + 1) − 1/(x + 2). Integrating the three pieces gives x + ln|x + 1| − ln|x + 2| + c. M1 for dividing before splitting, A1 for the quotient plus remainder form, M1 for partial fractions on the remainder, A1 for the two fractions, M1 for integrating, A1 for the answer. Going straight to partial fractions without dividing produces an identity that cannot be satisfied, and the resulting answer loses the whole of the x term.

Question 13

Find the exact value of ∫₀² (4x + 14)/(x² + 7x + 12) dx, giving your answer as a single logarithm.

Accepted Answer

Read the top against the derivative of the bottom before splitting anything. The bottom differentiates to 2x + 7, and the top is 4x + 14 = 2(2x + 7), so the f′/f pattern applies and the integral is 2 ln(x² + 7x + 12). No modulus is needed, because the bottom runs from 12 at the lower limit to 30 at the upper and stays positive throughout. Substituting the limits, 2(ln 30 − ln 12) = 2 ln(5/2) = ln(25/4). M1 for recognising the top as twice the derivative of the bottom, A1 for 2 ln(x² + 7x + 12), M1 for substituting both limits, A1 for 2(ln 30 − ln 12), M1 for compressing with the log laws, A1 for ln(25/4). Partial fractions are available here as well, since the bottom is (x + 3)(x + 4) and the split is 2/(x + 3) + 2/(x + 4), and the two logarithms recombine to the same value.

Question 14

Show that ∫1/(x² − a²) dx = (1/2a) ln|(x − a)/(x + a)| + c, where a is a positive constant, and hence find the exact value of ∫₃⁴ 1/(x² − 4) dx.

Accepted Answer

The bottom is a difference of two squares, so it factorises: x² − a² = (x − a)(x + a). Write 1/[(x − a)(x + a)] = A/(x − a) + B/(x + a) and multiply up, giving 1 = A(x + a) + B(x − a). Substituting x = a gives 1 = 2aA, and substituting x = −a gives 1 = −2aB. M1 for the partial-fraction set-up. So A = 1/(2a) and B = −1/(2a). A1 for both constants. The integrand is therefore (1/2a)[1/(x − a) − 1/(x + a)], and each piece is a logarithm of a linear bracket with inner coefficient 1: (1/2a)[ln|x − a| − ln|x + a|]. M1 for integrating both pieces. The subtraction law puts them together as (1/2a) ln|(x − a)/(x + a)| + c, as required. A1 for the printed form. Now take a = 2. Across [3, 4] both brackets are positive, so no modulus is needed: the value is (1/4)[ln((x − 2)/(x + 2))] from 3 to 4 = (1/4)[ln(2/6) − ln(1/5)]. M1 for a = 2 with both limits substituted. That is (1/4)[ln(1/3) + ln 5] = (1/4) ln(5/3). A1 for the exact value.

Question 15

A student begins to write (2x + 6)/(x² + 6x + 9) as A/(x + 3) + B/(x + 3)². Show that this fraction is not in its lowest terms, find ∫(2x + 6)/(x² + 6x + 9) dx, and state what the student's route would have produced.

Accepted Answer

Factorise both parts before anything else. The bottom is a perfect square, x² + 6x + 9 = (x + 3)², and the top is 2x + 6 = 2(x + 3). M1 for factorising both. The two share the factor x + 3, so the fraction is not in its lowest terms. A1 for the shared factor. Cancelling leaves 2/(x + 3), a constant over a single linear bracket, and its integral is 2 ln|x + 3| + c. A1 for the integral. The student's template is not wrong, only longer than it needs to be. Multiplying up gives 2x + 6 = A(x + 3) + B; substituting x = −3 gives B = 0, and comparing the coefficients of x then gives A = 2. B1 for B = 0 with A = 2. That route reaches 2/(x + 3) as well, and the same logarithm after it. B1 for cancelling first.