Integration Techniques: Partial fractions; IBP with \(\ln\) — RI 2025 H2 Math Prelim Paper 1

(a) Let \(\dfrac{9x}{(2x-1)(x+1)^2} = \dfrac{A}{2x-1}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2}\).

Then \(9x \equiv A(x+1)^2 + B(2x-1)(x+1) + C(2x-1)\).

Sub. \(x = \tfrac{1}{2}\): \(A = 2\). Sub. \(x = -1\): \(C = 3\).

Comparing coefficient of \(x^2\): \(A + 2B = 0 \Rightarrow B = -1\).

\[\begin{aligned} \int\frac{9x}{(2x-1)(x+1)^2}\,\mathrm{d}x &= \int\!\left(\frac{2}{2x-1}-\frac{1}{x+1}+\frac{3}{(x+1)^2}\right)\!\mathrm{d}x\\ &= \ln|2x-1| - \ln|x+1| - \frac{3}{x+1} + c\\ &= \ln\!\left|\frac{2x-1}{x+1}\right| - \frac{3}{x+1} + c, \quad c\in\mathbb{R} \end{aligned}\]

(b)(i) \[ \frac{\mathrm{d}}{\mathrm{d}x}\!\left(\frac{1}{x^2+1}\right) = \frac{-2x}{(x^2+1)^2} \]

(b)(ii) \[ \frac{\mathrm{d}}{\mathrm{d}x}\ln\sqrt{x^2+1} = \frac{\mathrm{d}}{\mathrm{d}x}\!\left[\tfrac{1}{2}\ln(x^2+1)\right] = \frac{x}{x^2+1} \]

(b)(iii) Integrate by parts with \(u = \ln\sqrt{x^2+1}\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{-2x}{(x^2+1)^2}\): \[ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{x}{x^2+1}, \qquad v = \frac{1}{x^2+1} \]

\[\begin{aligned} \int\frac{x\ln\sqrt{x^2+1}}{(x^2+1)^2}\,\mathrm{d}x &= -\frac{1}{2}\int\frac{-2x}{(x^2+1)^2}\ln\sqrt{x^2+1}\,\mathrm{d}x\\ &= -\frac{1}{2}\!\left[\frac{\ln\sqrt{x^2+1}}{x^2+1} - \int\frac{1}{x^2+1}\cdot\frac{x}{x^2+1}\,\mathrm{d}x\right]\\ &= -\frac{\ln\sqrt{x^2+1}}{2(x^2+1)} + \frac{1}{2}\int\frac{x}{(x^2+1)^2}\,\mathrm{d}x\\ &= -\frac{\ln\sqrt{x^2+1}}{2(x^2+1)} - \frac{1}{4(x^2+1)} + c\\ &= -\frac{\ln(x^2+1)+1}{4(x^2+1)} + c, \quad c\in\mathbb{R} \end{aligned}\]