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?

Let $u=\sqrt{2-3x}$, then $1+x^2 = \frac{1}{9}\left((2-u^2)^2+9\right)$ and $x=\frac{1}{3}\left(2-u^2\right)$ so $dx=-\frac{2}{3}u\,du$ so you want:
$$-6\int\frac{u^2\,du}{(2-u^2)^2+9}$$
Now attempting partial fractions, we get a factoring:
$$\begin{align}(2-u^2)^2+9 &= u^4-4u^2+13\\&=\left(u^2+\left(\sqrt{4+2\sqrt{13}}\right)u+\sqrt{13}\right)\left(u^2-\left(\sqrt{4+2\sqrt{13}}\right)u+\sqrt{13}\right)
\end{align}$$
That's gonna involve some very ugly formula to get a final result. I'd be surprised if there is a clean formula.
If $\alpha = \sqrt{1+\sqrt{\frac{13}{4}}}$ and $\beta = \sqrt{13}$, then you need to solve:
$$f(u)=\frac{u^2}{u^4-4u^2+13}=\frac{au+b}{u^2+2\alpha u +\beta}+\frac{cu+d}{u^2-2\alpha u +\beta}$$
Since $f(u)=f(-u)$, you have that $-c=a,b=d$. Since $b+d=0$, you get $b,d=0$. 
You get $$f(u)=\frac{au}{u^2+2\alpha u+\beta}+\frac{-au}{u^2-2\alpha u+\beta}$$
And you get $-4a\alpha =1$ of $a=-\frac{1}{4\alpha}$.
You are going to get some horrible mix of $\arctan, \log$ and lots of square roots of $13$.
Finally, $$\begin{align}\int\frac{u\,du}{u^2+2\alpha u+\beta} &= \frac{1}{2}\int \frac{(2u+2\alpha)\,du}{u^2+2\alpha u+\beta}-\alpha\int\frac{du}{u^2+2\alpha u+\beta}
\\&=\frac{1}{2}\log(u^2+2\alpha u+\beta)+\frac{\sqrt{\beta-\alpha^2}}\arctan\left(\frac{u+\alpha}{\sqrt{\beta-\alpha^2}}\right)
\end{align}$$
And likewise:
$$\int\frac{u\,du}{u^2-2\alpha u+\beta}=\frac{1}{2}\log(u^2-2\alpha u+\beta)+\frac 1{\sqrt{\beta-\alpha^2}}\arctan\left(\frac{u-\alpha}{\sqrt{\beta-\alpha^2}}\right)$$
Finally, you get, with $\gamma = \sqrt{\beta-\alpha^2}=\sqrt{-1+\sqrt{\frac{13}4}}$:
$$\frac{3}{2\alpha}\left(\frac{1}{2}\log\frac{u^2+2\alpha u+\beta}{u^2-2\alpha u+\beta}+\frac1\gamma\left(\arctan\left(\frac{u+\alpha}{\gamma}\right)-\arctan\left(\frac{u-\alpha}{\gamma}\right)\right)\right)$$
As it turns out, $\alpha\gamma = \frac{3}{2}$, so you get some simplifications. You can also use that $\arctan x-\arctan y = \arctan\frac{x-y}{1+xy}.$
Then replace $u=\sqrt{3-2x}$.
