I have to integrate $$\int \frac{x^3 + 1}{8x + \sqrt{9-x^2}}dx$$
in a rational way.
So as I understand I must find such "replacements" (I mean $u$, $du$ etc.) that would transform this function into a rational function. But I am getting frustrated while doing this, maybe because I am picking wrong variables for my $du$, I don't know..
One possible way would be this: $$\frac{x^3+1}{8x+\sqrt{9-x^2}}=\frac{(x^3+1)(8x-\sqrt{9-x^2})}{(8x)^2-(\sqrt{9-x^2})^2}=\frac{8x^4-x^3\sqrt{9-x^2}+8x-\sqrt{9-x^2}}{65x^2-9}$$ and you can factor the bottom as: $$65x^2-9=(\sqrt{65}x+3)(\sqrt{65}x-3)$$ and split the top up and make multiple substitutions, however you still have a fraction involving a squareroot which isn't always ideal.
Alternatively, consider the substitution $x=3\sin(u)$, this converts your integral into: $$\frac{27\sin^3u+1}{24\sin u+\cos u}$$ then you can use the fact that: $$24\sin u+\cos u=\sqrt{577}\sin(u+\alpha)$$ where $\tan\alpha=\frac1{24}$ then make the substitution $v=u+\alpha$