Integration techniques: TJC 2024 Promo Q4

(a)

$$\begin{aligned} \frac{x^2+4}{x^2-2x+5} &=1+\frac{2x-1}{x^2-2x+5}\\ &=1+\frac{2x-2}{x^2-2x+5}+\frac{1}{x^2-2x+5} \end{aligned}$$ $$\begin{aligned} \int\frac{x^2+4}{x^2-2x+5}\,\mathrm{d} x &=\int 1+\frac{2x-2}{x^2-2x+5}+\frac{1}{(x-1)^2+2^2}\,\mathrm{d} x\\ &=x+\ln(x^2-2x+5)+\frac12\tan^{-1}\!\left(\frac{x-1}{2}\right)+C \end{aligned}$$

(Q4b)

$$\begin{aligned} \frac{\,\mathrm{d}}{\,\mathrm{d} x}\sec^2 3x &=2(\sec3x)(\sec3x\tan3x)(3)\\ &=6\tan3x\sec^2 3x \end{aligned}$$ $$\begin{aligned} \int x\tan3x\sec^2 3x\,\mathrm{d} x &=\frac16\int x(6\tan3x\sec^2 3x)\,\mathrm{d} x\\ &=\frac16\left[x\sec^2 3x-\int1(\sec^2 3x)\,\mathrm{d} x\right]\\ &=\frac16\left(x\sec^2 3x-\frac13\tan3x\right)+C \end{aligned}$$