The value of the integral $$ \frac{d}{d x}\left[\int_{3}^{e^{x}}\left(t^{2}+\sqrt{t}+1\right) d t\right]= $$

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To find the derivative of the integral

$$
\frac{d}{dx}\left[\int_{3}^{e^{x}}\left(t^{2}+\sqrt{t}+1\right) dt\right],
$$

we can apply the **Leibniz rule** for differentiation under the integral sign. The Leibniz rule states:

$$
\frac{d}{dx} \left[ \int_{a(x)}^{b(x)} f(t) \, dt \right] = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).
$$

In this case:
- $$ a(x) = 3 $$ (a constant), so $$ a'(x) = 0 $$.
- $$ b(x) = e^x $$, so $$ b'(x) = e^x $$.
- The integrand $$ f(t) = t^2 + \sqrt{t} + 1 $$.

Applying the Leibniz rule:

$$
\frac{d}{dx} \left[\int_{3}^{e^{x}}\left(t^{2}+\sqrt{t}+1\right) dt\right] = \left( (e^x)^2 + \sqrt{e^x} + 1 \right) \cdot e^x.
$$

Simplifying the expression inside the parentheses:

$$
(e^x)^2 = e^{2x}, \quad \sqrt{e^x} = e^{x/2}.
$$

So, the derivative becomes:

$$
e^{2x} \cdot e^x + e^{x/2} \cdot e^x + 1 \cdot e^x = e^{3x} + e^{3x/2} + e^x.
$$

**Final Answer:**

After simplifying, the derivative is e³ˣ plus e³⁄²ˣ plus eˣ. Thus,

e^{3x} + e^{3x/2} + e^{x}
The derivative is $$ e^{3x} + e^{3x/2} + e^{x} $$.

The value of the integral

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