Integration techniques: DHS 2024 Promo Q10

(a) $$\begin{aligned} \int\frac{1-3x}{1+9x^2}\,dx &=\int\frac1{1+9x^2}\,dx-\int\frac{3x}{1+9x^2}\,dx\\ &=\frac19\int\frac1{(\frac13)^2+x^2}\,dx-\frac16\int\frac{18x}{1+9x^2}\,dx\\ &=\frac13\tan^{-1}(3x)-\frac16\ln(1+9x^2)+c \end{aligned}$$ (b) $x=2\sin\theta\Rightarrow\dfrac{dx}{d\theta}=2\cos\theta$ $$\begin{aligned} \int\frac{(x-1)^2}{\sqrt{4-x^2}}\,dx &=\int\frac{(2\sin\theta-1)^2}{\sqrt{4-4\sin^2\theta}}(2\cos\theta)\,d\theta\\ &=\int\frac{(2\sin\theta-1)^2}{2\cos\theta}(2\cos\theta)\,d\theta\\ &=\int4\sin^2\theta-4\sin\theta+1\,d\theta\\ &=\int4\left(\frac{1-\cos2\theta}{2}\right)-4\sin\theta+1\,d\theta\\ &=\int3-2\cos2\theta-4\sin\theta\,d\theta\\ &=3\theta-\sin2\theta+4\cos\theta+c\\ &=3\sin^{-1}\left(\frac x2\right)-\frac{x\sqrt{4-x^2}}2+2\sqrt{4-x^2}+c \end{aligned}$$ (c) $$\begin{aligned} \frac d{dx}\left(\sin^{-1}(2x^2)\right) &=\frac{4x}{\sqrt{1-(2x^2)^2}}\\ &=\frac{4x}{\sqrt{1-4x^4}} \end{aligned}$$ $$\begin{aligned} \int\left(2x\sin^{-1}(2x^2)\right)dx &=x^2\sin^{-1}(2x^2)-\int\frac{4x^3}{\sqrt{1-4x^4}}\,dx\\ &=x^2\sin^{-1}(2x^2)+\frac14\int\frac{-16x^3}{\sqrt{1-4x^4}}\,dx\\ &=x^2\sin^{-1}(2x^2)+\frac14\frac{(1-4x^4)^{1/2}}{1/2}+c\\ &=x^2\sin^{-1}(2x^2)+\frac12(1-4x^4)^{1/2}+c \end{aligned}$$