Substitution needed for calculating integral $\int \frac{1}{x \sqrt{x^4+x^2+1}}dx$

Question

Can you find two functions: $\phi:(0,\infty) \longrightarrow (0,\infty) $, $f:(0,\infty)\longrightarrow \mathbb{R}$, with $\phi$ differentiable, such that $f(\phi(x))\phi'(x)=\frac{1}{x \sqrt{x^4+x^2+1}},\forall x \in \mathbb{R}$?

I want to calculate the integral $$\int \frac{1}{x \sqrt{x^4+x^2+1}}dx$$ without to use a trigonometric substitution, such that the integral reduces to $\int f(t)dt$, which I hope is easier to calculate. I don't want any method which has nothing to do with this form. Thank you!

Answer 1

I think this question is a bit strange; one should use the best method available, and solve such an integral step by step getting closer to the solution. In case you just want the answer, though: $$ f(u)=\frac{2}{u^2-4} $$ and $$ \phi(x)=\sqrt{3}+\frac{2\sqrt{1+x^2+x^4}-\sqrt{3}}{1+2x^2}, $$ but I'm almost ashamed to tell...

(I admit that $f$ is not defined at $u=2$, but since $\phi(x)>2$ for $x>0$, that is not a real problem.)

Answer 2

Per the Euler substitution $t=\sqrt{x^4+x^2+1}-x^2\in(0,1]$ $$\int \frac{dx}{x\sqrt{x^4+x^2+1}}=\int \frac{dt}{t^2-1}=-\tanh^{-1}t+C $$