2022 H2 Mathematics Paper 1 Question 8

Definite Integrals: Areas and Volumes

Answers

(a)

\ln (x^2 + 2 x + 1) + \frac{3}{x+1} + c.

(b)

\ln \frac{16}{9}.

Full solutions

(a)

\begin{align*} & \int \frac{2x-1}{x^2 + 2 x + 1} \mathop{\mathrm{d}x} \\ &= \int \frac{2x+2}{x^2 + 2 x + 1} \mathop{\mathrm{d}x} - \int \frac{3}{x^2 + 2 x + 1}\mathop{\mathrm{d}x} \\ &= \ln (x^2 + 2 x + 1) - \int \frac{3}{(x + 1)^2} \mathop{\mathrm{d}x} \\ &= \ln (x^2 + 2 x + 1) + \frac{3}{x + 1} + c \; \blacksquare \end{align*}

(b)

\begin{align*} & \int_0^2 \frac{|2x-1|}{x^2 + 2 x + 1} \mathop{\mathrm{d}x} \\ &= - \int_0^{\frac{1}{2}} \frac{2x-1}{x^2 + 2 x + 1} \mathop{\mathrm{d}x} + \int_{\frac{1}{2}}^2 \frac{2x-1}{x^2 + 2 x + 1} \mathop{\mathrm{d}x} \\ &= - \left[ \ln (x^2 + 2 x + 1) + \frac{3}{x + 1} \right]_0^{\frac{1}{2}} \\ & \qquad \quad + \left[ \ln (x^2 + 2 x + 1) + \frac{3}{x + 1} \right]_{\frac{1}{2}}^2 \\ &= - \left( \ln \frac{9}{4} + \frac{3}{\frac{3}{2}} - \ln 1 - 3 \right) + \left( \ln 9 + \frac{3}{3} - \ln \frac{9}{4} - \frac{3}{\frac{3}{2}} \right) \\ &= - \ln \frac{9}{4} + 1 + \ln 9 - \ln \frac{9}{4} - 1 \\ &= \ln \frac{16}{9} \; \blacksquare \end{align*}

Question Commentary

My typical recommendation for all integration techniques question: for the first step check if the ${f'(x)}$ formulas. For part (a) of the question they almost do, so the trick for the question is to "force" ${f'(x)}$ via addition/subtraction and splitting up into two integrals. The second integral is then done via factorization/completing the square.

Part (b) tests on the integration of modulus functions. It doesn't come out that often (last seen in 2011 paper 1 question 5), but it's definitely in most notes/tutorials, where the trick is to split up the integral based on when the function within the modulus is positive/negative. Thereafter don't forget to use the result from part (a).