Find the following integrals. \(\displaystyle\int \frac{x^2 - 3x - 1}{x^2 - 4x + 1} \, \mathrm{d}x \

Question

Find the following integrals. $\displaystyle\int \frac{x^2 - 3x - 1}{x^2 - 4x + 1} \, \mathrm{d}x \,,$[3] $\displaystyle\int x \ln (kx) \, \mathrm{d}x \,,$ where $k$ is a real constant,[2] $\displaystyle\int \frac{\sin \theta}{2 \cos 2\theta + 1} \, \mathrm{d}\theta \,,$ using the substitution $x = \cos \theta$.[4]

Accepted Answer

(a)$x+\frac{1}{2}\ln |x^2-4x+1|+C$ (b)$\frac{1}{2}x^2\ln kx -\frac{1}{4}x^2+C$ (c)$\frac{1}{4}\ln|\frac{1+2\cos\theta}{1-2\cos\theta}|+C$

2025 TMJC P1 Q3

2025 TMJC P1 Q3