<s>[INST] Find $\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$ Find $$\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$$
I used substitution $x=\frac{2}{3} \sin^2 y$ So we get $dx=\frac{4}{3} \sin y \cos y dy$
hence
$$I= \frac{36\sqrt{2}}{3} \int \frac{\sin y \cos^2 y\, dy}{4 \sin^4 y+9}$$
Now if again i use substitution $\cos y=t$ we get
$$I=-12\sqrt{2} \int \frac{t^2 \,dt}{4t^4-8t^2+13}$$
$$I=-12 \sqrt{2} \int \frac{dt}{4t^2+\frac{13}{t^2}-8}$$
Any other approach?
 [/INST] hint: let $u = \sqrt{2-3x}\implies 2-3x=u^2\implies x = \dfrac{2-u^2}{3}, dx = -\dfrac{2udu}{3}$. Then do a fraction decomposition. It can be done but its not short !
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