$\int\frac{1}{x\sqrt{x^4+x^2+1}}dx$

Question

Calculate $$\int\frac{1}{x\sqrt{x^4+x^2+1}}dx.$$ I applied the formula that says : $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$ but I do not think it helps me in some way.

Answer 1

Prompt: $$ \begin{align*} \int{\frac{1}{x\sqrt{x^4+x^2+1}}\text{d}x}&=\int{\frac{\frac{1}{x^2}}{\sqrt{x^2+\frac{1}{x^2}+1}}\text{d}x} \\ &=\frac{1}{2}\int{\frac{\frac{1}{x^2}-1}{\sqrt{x^2+\frac{1}{x^2}+1}}\text{d}x}+\frac{1}{2}\int{\frac{1+\frac{1}{x^2}}{\sqrt{x^2+\frac{1}{x^2}+1}}\text{d}x} \\ &=-\frac{1}{2}\int{\frac{1}{\sqrt{\left( x+\frac{1}{x} \right) ^2-1}}\text{d}\left( x+\frac{1}{x} \right)}+\frac{1}{2}\int{\frac{1}{\sqrt{\left( x-\frac{1}{x} \right) ^2+3}}\text{d}\left( x-\frac{1}{x} \right)} \\ &=… \end{align*} $$