Proving $\int_a^bf(x)^2dx=F(b)F'(b)-F(a)F'(a)-\int_a^bF(x)F''(x)dx$, where $F'(x)=f(x)$ (without using integration by parts) 
Let $f\colon [a, b] \to \mathbb{R}$ be a differentiable function such that $F$ is an antiderivative of $f$ in $[a, b]$. Without using integration by parts, prove that
$$\int_a^b f(x)^2 \, dx = F(b)F' (b) - F(a)F'(a) -\int_a^b F(x)F''(x) \, dx$$
Suggestion: $$(f \cdot g)'= f' \cdot g + f \cdot g'$$

I used the Fundamental Theorem of Calculus and Barrow's rule, after that I don't know how to continue.
I did was $$\frac{d}{dx} F(x) \cdot F'(x) = f^2(x) + F(x) \cdot F''(x)$$

As you said
$$(FF')' = (F')^2 + FF'' \implies$$
$$(FF')' = f^2 + FF'' \implies$$
$$f^2 = (FF')' - FF'' \implies$$
$$\int_a^b f^2 = \int_a^b(FF')' - \int_a^bFF'' \implies$$
$$\int_a^bf^2 = \Bigl[FF'\Bigr]_a^b-\int_a^bFF''$$
