2014 H2 Mathematics Paper 2 Question 2

Integration Techniques

Answers

32ln⁡1345+83tan⁡−123{\frac{3}{2} \ln \frac{13}{45} + \frac{8}{3} \tan^{-1} \frac{2}{3}}

Full solutions

9x2+x−13(2x−5)(x2+9)=A2x−5+Bx+Cx2+9\frac{9 x^2 + x - 13}{(2 x - 5)(x^2 + 9)} = \frac{A}{2 x - 5} + \frac{Bx+C}{x^2 + 9}
By the cover up rule,
A=9(52)2+(52)−13(52)2+9=3\begin{align*} A &= \frac{9\left( \frac{5}{2} \right)^2+\left( \frac{5}{2} \right)-13}{\left( \frac{5}{2} \right)^2+9} \\ &= 3 \end{align*}
9x2+x−13=3(x2+9)+(Bx+C)(2x−5)9 x^2 + x - 13 = 3(x^2 + 9) + (Bx+C)(2 x - 5)
Comparing coefficients,
x2:2B+3=9B=3x0:−5C+27=−13C=8\begin{align*} &x^2:& \quad 2B + 3 &= 9 \\ && B = 3 \\ &x^0:& -5C + 27 &= -13 \\ && C = 8 \end{align*}
∫029x2+x−13(2x−5)(x2+9)  dx=∫0232x−5+3x+8x2+9  dx=∫0232x−5+3xx2+9+8x2+9  dx=[32ln⁡∣2x−5∣+32ln⁡(x2+9)+83tan⁡−1x3]02=32(ln⁡1−ln⁡5)+32(ln⁡13−ln⁡9)+83tan⁡−123=32(ln⁡13−ln⁡5−ln⁡9)+83tan⁡−123=32ln⁡1345+83tan⁡−123  ■\begin{align*} & \int_0^2 \frac{9 x^2 + x - 13}{(2 x - 5)(x^2 + 9)} \; \mathrm{d}x \\ & = \int_0^2 \frac{3}{2 x - 5} + \frac{3x+8}{x^2 + 9} \; \mathrm{d}x \\ & = \int_0^2 \frac{3}{2 x - 5} + \frac{3x}{x^2 + 9} + \frac{8}{x^2 + 9} \; \mathrm{d}x \\ & = \left[ \frac{3}{2} \ln \left| 2 x - 5\right| + \frac{3}{2}\ln \left( x^2 + 9 \right) + \frac{8}{3}\tan^{-1} \frac{x}{3} \right]_0^2 \\ & = \frac{3}{2} \left( \ln 1 - \ln 5 \right) + \frac{3}{2} \left( \ln 13 - \ln 9 \right) + \frac{8}{3} \tan^{-1} \frac{2}{3} \\ & = \frac{3}{2} \left( \ln 13 - \ln 5 - \ln 9 \right) + \frac{8}{3} \tan^{-1} \frac{2}{3} \\ & = \frac{3}{2} \ln \frac{13}{45} + \frac{8}{3} \tan^{-1} \frac{2}{3} \; \blacksquare \end{align*}