Proving. By integration

Question

If F(x) = integral (from 1 to x) [f(t)]dt , where f(t) = integral (from 1 to t^2) [(sqrt1+u^2)/u]du, find F"(2)

Answer 1

F(x) = integral (from 1 to x) f(t)dt

By the Fundamental Theorem of Calculus(FTC), F'(x) = f(x). So, F"(x) = f'(x).

But, f(t) = integral (from 1 to t^2) [sqrt(1+u^2) / u]du

Again, using the FTC, f'(t) = (sqrt[1 + (t^2)^2] / t^2) (t^2)' = 2 sqrt[1 + t^4] / t

So, F"(2) = f'(2) = 2sqrt(17) / 2 = sqrt(17).