confusion in using Lebiniz integral rule I was trying this question - 
 Let $$f: (0,\infty )\rightarrow \mathbb{R}$$ and $$F(x) =
 \int_{0}^{x}tf(t)dt$$ If $F(x^2)= x^{4} + x^{5} $, then the value of $\sum_{r=1}^{12}f(r^{2})$ is
I applied chain rule for differentiation in $F(x^2)$ to get $$2xF'(x^2)=4x^3+5x^4$$ then used Leibniz rule in $F(x^2)$ to get 
$$F'(x^2)=2x(x^2)f(x^2)$$
 and substituted it in above equation to get $f(r^2)$ then trying to sum it up but in the solution the correct equation is $ 2x(x^2)f(x^2)= 4x^3 + 5x^4 $ but according to my method the equation for $f(r^2)$ is $2x(2x)(x^2)f(x^2)= 4x^3 + 5x^4$, what am I doing wrong?

Since $F(x^2)=x^4+x^5=(x^2)^2+(x^2)^{5/2}$, we have $F(x)=x^2+x^{5/2}$, and $f(x)=x^{-1}F'(x)=x^{-1}(2x+\frac{5}{2}x^{3/2})=2+\frac52\sqrt{x}$. It follows that  $f(x^2)=2+\frac52\sqrt{x^2}=2+\frac52|x|$. Hence
$$
\sum_{r=1}^{12}f(r^2)=\sum_{r=1}^{12}(2+\frac52r)=2\cdot12+\frac52\cdot\frac{12(12+1)}{2}=24+195=219.
$$
