If $$m$$ and $$n$$ respectively are the number of local maximum and local minimum points of the function $$f(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} dt$$, then the ordered pair $$(m, n)$$ is equal to

Given,

$$f(x)=\int_0^{x^2}\frac{t^2-5t+4}{2+e^t}\,dt$$

Using Leibnitz rule,

$$f'(x)=\frac{x^4-5x^2+4}{2+e^{x^2}}\cdot2x$$

$$=\frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}$$

Since,

$$2+e^{x^2}>0$$

for all $$x$$, the critical points are obtained from

$$2x(x^2-1)(x^2-4)=0$$

Hence,

$$x=0,\pm1,\pm2$$

Now, checking sign changes of

$$x(x^2-1)(x^2-4)$$

we get:

- Increasing on

$$(-\infty,-2),\ (-1,0),\ (1,2)$$

- Decreasing on

$$(-2,-1),\ (0,1),\ (2,\infty)$$

Therefore,

- Local maxima occur at

$$x=-2,\ 0,\ 2$$

So,

$$m=3$$

- Local minima occur at

$$x=-1,\ 1$$

So,

$$n=2$$

Hence,

$$\boxed{(m,n)=(3,2)}$$