If $F(x) = \int_{1}^{\,x}{f(t)}\,dt$, where $f(t) = \int_{1}^{\,t^2}{\frac{\sqrt{9+u^4}}{u}}\,du$, find $F''(2).$ If $\displaystyle F(x) = \int_{1}^{\,x}{f(t)}\,dt$, where $\displaystyle f(t) = \int_{1}^{\,t^2}{\frac{\sqrt{9+u^4}}{u}}\,du$, find $F''(2).$
I used FTC to get
If $\displaystyle F(x) =\int_{1}^{\,x}{\frac{\sqrt{9+x^8}}{x}}\,dx $
Then I tried to use FTC again to find $F'(x)$ but then I got lost cause it's just the same thing over again. So then I decided that i'll just plug in 2 to the function and got 8.14 but I know this is incorrect. Any ideas?

The FTC says that $F'(x)=f(x)=\int_{1}^{x^2}{\frac{\sqrt{9+u^4}}{u}}\,du$. Now use the FTC again along with the chain rule. To do that note that $f(x)=g(h(x))$ where $g(x):=\int_{1}^{x}{\frac{\sqrt{9+u^4}}{u}}\,du$ and $h(x):=x^2$. Hence $F''(x)=f'(x)=g'(h(x))h'(x)=\frac{\sqrt{9+(x^2)^4}}{x^2}\cdot2x$. Evaluating at $2$ gives $\sqrt{265}$.
