Calculating the Derivative of an Integral Function

Question:

If $F(x) = \int_{1}^{2x} \frac{1}{1 - t^3} \, dt$ , then $F'(x) =$

Understand the Function Structure:
Given the function:

$F(x) = \int_{1}^{2x} \frac{1}{1 - t^3} \, dt$

Here, the upper limit of integration is $g(x) = 2x$ .
Apply the Fundamental Theorem of Calculus:
If $F(x) = \int_{a}^{g(x)} f(t) \, dt$ , then:

$F'(x) = f(g(x)) \cdot g'(x)$

Where:
- $f(t) = \frac{1}{1 - t^3}$
- $g(x) = 2x$
- $g'(x) = 2$
Compute $F'(x)$ :
$F'(x) = \frac{1}{1 - (2x)^3} \cdot 2 = \frac{2}{1 - 8x^3}$
Match with the Given Options:
The derivative simplifies to:

$F'(x) = \frac{2}{1 - 8x^3}$

This corresponds to option (E).