To Evaluate $$I=\int_{-1}^{1} \frac{dx}{x^2+x+1+\sqrt{x^4+3x^2+1}}$$
I rationalized the denominator getting as
$$I=\int_{-1}^{1}\frac{x^2+x+1-\sqrt{x^4+3x^2+1}}{2x(x^2+1)}$$ $\implies$
$$I=\int_{-1}^{1}\frac{dx}{2x}+\int_{-1}^{1}\frac{dx}{2(x^2+1)}+\int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}\:dx}{2x(x^2+1)}$$
First and Third integrals are zero as the Integrands are odd functions and hence
$$I=\int_{-1}^{1}\frac{dx}{2(x^2+1)}=\frac{\pi}{4}$$
No.
Your final result is right, but we have $$ \int_{-1}^{1}\frac{dx}{x}\ne0,\qquad \int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\ne0, \tag1 $$ each integral being divergent, it is the difference of the preceding integrals which vanishes, that is $$ \int_{-1}^{1}\left(\frac{1}{x}-\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\right)dx=0. \tag2 $$To prove $(2)$, one may observe the integrand is an odd function over $[-1,1]$ and one may observe the integral is convergent, since the singularity of the integrand as $x \to 0$ is removable, $$ \left(\frac{1}{x}-\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\right) \sim -\frac x2, $$ the function being continuous elsewhere.