Given $f\left(x\right)={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$

We know that,

$\frac{d}{dx}{\int }_{u\left(x\right)}^{v\left(x\right)}f\left(v\right)dx=f\left(v\left(x\right)\right)\frac{d}{dx}v\left(u\right)-f\left(u\left(x\right)\right)\frac{d}{dx}u\left(x\right)$

Using above formula we get $f^{\text{'}}\left(x\right)=\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)·2x$

For critical points $f^{\text{'}}\left(x\right)=0$

So, $\frac{x\left(x^2-1\right)\left(x^2-4\right)}{2+e^{x^2}}=0$

$\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)=0$

Now by first derivative test we get,

Here points of maxima are $\left\{-1,1\right\}$  & points of minima are $\left\{-2,0,2\right\}$