If $F(x) = \int_{1}^{x^2} (\sqrt{1 + u}) du$, find $F'(x)$

Question

I am not sure how to use the second fundamental theorem of calculus to solve this, as the upper bound for this integral is $x^2$ and not $x$.

Answer 1

Chain Rule: $\dfrac{d}{d(x^{2})}\displaystyle\int_{1}^{x^{2}}\sqrt{1+u}du\cdot\dfrac{d(x^{2})}{dx}$.

Answer 2

Hint

If $G(x)=\int_1^x h(x)du$ then, by definition, $G'(x)=h(u)$. Here, you have that $$F(x)=G(x^2)$$ with $h(u)=\sqrt{1+u}$.