Find the integral by first principle In my math course homework, I encountered this problem:

Find $$\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}$$ by first principle

$$\begin{split}\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_0\frac{dt}{1+e^{t^2}}-\int^{x^2}_0\frac{dt}{1+e^{t^2}}}{h}\\&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}\end{split}$$
Then I don't know how to do. Please help me!

Let,
$$F(x)=\int_{0}^{x^2}\frac{dt}{1+e^{t^{2}}}$$
$$\Rightarrow F(\sqrt{x})=\int_{0}^{x}\frac{dt}{1+e^{t^{2}}}$$
Let,
$$f(x)=\frac{1}{1+e^{x^{2}}}$$
From the first principle,
$$ \frac{dF(\sqrt{x})}{dx}=f(x)$$
$$\Rightarrow \frac{F'(\sqrt{x})}{2\sqrt{x}}=\frac{1}{1+e^{x^{2}}}$$
$$\Rightarrow {F'(\sqrt{x})}=\frac{{2\sqrt{x}}}{1+e^{x^{2}}}$$
$$\Rightarrow {F'({x})}=\frac{{2{x}}}{1+e^{x^{4}}}$$
