<s>[INST] Calculating $H'(x)$ given $H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$ Given $\displaystyle H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$, we want to find $H'(x)$. 
First, we rewrite $H(x)$ as follows:
$$\begin{align}
&= \int_0^{x^2 + 2x} e^{-t^2} dt + \int_{x^3 + 1}^0 e^{-t^2} dt \qquad &\text{Properties of integrals} \\
&= \int_0^{x^2 + 2x} e^{-t^2} dt - \int_{0}^{x^3 + 1} e^{-t^2} dt \qquad &\text{Definition of backwards integrals} \tag{1}
\end{align}
$$
Next, we'll define $\displaystyle F(x) = \int_0^x e^{-t^2} dt$.
We know its derivative is $F'(x) = e^{-x^2}$, by the Fundamental Theorem of Calculus.
Next, we'll define new functions for the two integrals in $(1)$:
$$\begin{align*}
H_1(x) &= \int_0^{x^2 + 2x} e^{-t^2} dt  &\qquad H_2(x) &= \displaystyle\int_{0}^{x^3 + 1} e^{-t^2} dt \\
&= F(x^2 + 2x)& &=F(x^3 + 1)
\end{align*}$$ 
We use the chain rule to find their derivatives:
$$ H_1'(x) = e^{-(x^2 + 2x)^2} (2x + 2) \qquad H_2'(x) = e^{-(x^3 + 1)^2} (3x) $$
Therefore,
$$H'(x) = e^{-(x^2 + 2x)^2} (2x + 2) - e^{-(x^3 + 1)^2} (3x)$$
Is my calculation correct?
 [/INST] Use Lebnitz rule:
$$\frac{d}{dx} \int_{L(x)}^{U(x)} f(t) dt= U'(x) f(U(x)) -L'(x) f(L(x))$$
So in your case you get $$H'(x)=(2x+2) e^{-(x^2+2x)^2}-3x^2 e^{-(x^3+1)^2}.$$
</s>