Express $\frac{1-6x}{(x+2)(3x^2+1)}$ in the form of $\frac{a}{x+2}+\frac{bx+c}{3x^2+1}$, where&n

Question

Express $\frac{1-6x}{(x+2)(3x^2+1)}$ in the form of $\frac{a}{x+2}+\frac{bx+c}{3x^2+1}$, where $a$, $b$ and $c$ are integers.[4] Differentiate $\ln(3x^2+2)$ with respect to $x$.[2] Using your answer to parts (a) and (b), show that $\int_{0}^{2} \frac{1-6x}{(x+2)(3x^2+1)} \, \mathrm{d}x = \ln 2 - \frac{1}{2} \ln 13.$[4]

Accepted Answer

(a)a=1, b=-3, c=0 (b)$\frac{6x}{3x^2+1}$

2025 MGS PRELIMS P1 Q8

2025 MGS PRELIMS P1 Q8