${\int }_0^1\frac{(1-x^2)dx}{x^4+x^2+1}=$

${\int }_0^1\frac{(1-x^2)dx}{x^4+x^2+1}$

$$\begin{gathered} ={\int }_0^1\frac{\left({\displaystyle \frac{1}{x^2}}-1\right)}{x^2+1+{\displaystyle \frac{1}{x^2}}}dx \\ Put x+\frac{1}{x}=t \end{gathered}$$

$$\begin{gathered} \left(1-\frac{1}{x^2}\right)dx=dt \\ and t^2={\left(x+\frac{1}{x}\right)}^2 \\ =x^2+\frac{1}{x^2}+2 \\ {t}^2-1=x^2+\frac{1}{x^2}+1 \\ x=0\Rightarrow t\to \infty \\ x=1\to t=1+\frac{1}{1}=2 \\ I=\underset{\infty }{\overset{2}{\int }}\frac{-dt}{t^2-1} \\ =\underset{2}{\overset{\infty }{\int }}\frac{dt}{t^2-1^2} \\ ={\left[\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|\right]}_2^{\infty } \\ =\frac{1}{2}\left[\log 1-\log \frac{1}{3}\right]=\frac{1}{2}\left[0+\log 3\right] \\ =\frac{1}{2}\log 3 \\ \left[\therefore \log \left|\frac{t-1}{t+1}\right|\right]=\log \left|\frac{1-{\displaystyle \frac{1}{t}}}{1+{\displaystyle \frac{1}{t}}}\right| \\ t\to \infty \Rightarrow \log \left|\frac{1-0}{1+0}\right|=\log 1=0] \end{gathered}$$