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!

It might be easier to look a little more abstractly and then apply to your specific case.
Suppose $f$ is continuous and $g$ is differentiable.
Let $\phi(x) = \int_0^{g(x)} f(t) dt$. The fundamental theorem/Leibniz rule tells us that
$\phi'(x) = f(g(x)) g'(x)$.
So we want to show that $\lim_{h \to 0} |{ \phi(x+h)-\phi(x) \over h} -f(g(x))g'(x)| = 0$.
Pick $\epsilon>0$ and $\delta_1 >0$ such that if $|t-g(x)| < \delta$ then $|f(t)-f(g(x))| < \epsilon$.
Also, choose $\delta_2 \le \delta_1$ such that if $|h|< \delta_2$ then
$|{g(x+h)-g(x) \over h} - g'(x)| < \epsilon$.
\begin{eqnarray}
|{\phi(x+h)-\phi(x) \over h } - f(g(x))g'(x)  | &=& |{1 \over h}\int_{g(x)}^{g(x+h)} f(t) dt - f(g(x))g'(x)| \\&=&
|{1 \over h}\int_{g(x)}^{g(x+h)} (f(g(x))+f(t)-(g(x))) dt - f(g(x))g'(x)| \\
&\le& |{1 \over h}\int_{g(x)}^{g(x+h)} f(g(x)) dt - f(g(x))g'(x)| +|{1 \over h}\int_{g(x)}^{g(x+h)}  \epsilon dt|\\
&=& |f(g(x))| | {(g(x+h)-g(x)) \over h} - g'(x) | + \epsilon |{g(x+h)-g(x) \over h} |\\
&\le& \epsilon |f(g(x))|+ \epsilon^2
\end{eqnarray}
It follows that the limit is zero.
