Find maximum on $[0,1]$ of $f(x) =\int_0^x \sqrt{t^4+(1-x)^2(x-t)^2} \operatorname d t$ This problem is taken from an exam to acced to PhD at my University in 2012.

Find the maximum on $[0,1]$ of the function defined by 
$$\tag{1}f(x) =\int_0^x \sqrt{t^4+(1-x)^2(x-t)^2}\,  \mathrm d t$$

Clearly $f(x) \ge 0$  for every $x \in [0,1]$, $f(0)=0$ and $f(1) = \frac{1}{3}$.

My attempt
My first idea was to eliminate the $x$ on the extrema, so with substitution $u = \frac{t}{x}$, we get
$$
f(x) = x \int_0^1 \sqrt{ x^2u^4 + (1-x)^2(1-u)^2}\,  \mathrm d u
$$
But this seems not  very promising.
Another idea that came to my mind was to use the Leibniz integral formula in formula $(1)$:
$$
f'(x) = x^2 + \int_0^x \frac{\partial }{\partial x} \sqrt{t^4+(1-x)^2(x-t)^2} \operatorname \, \mathrm d t
$$
But I  believe there is a simpler way, which I cannot find.
Could you provide me any help?
 A: I believe that your idea to use the Leibniz integral formula was exactly right and also simpler than you may have realized. 
Proceeding with that idea, we have:
$$f'(x) = x^2 + \int_0^x \frac{(1-x)(x-t)(1+t)}{\sqrt{t^4 + (1-x)^2(x-t)^2}} \, dt.$$
The integrand should always be non-negative for any $x \in [0,1]$. Hence, the integral is guaranteed to be non-negative. Similarly, $x^2$ is non-negative, so $f'(x)$ is non-negative. This guarantees that, for all $x \in [0,1]$, $f(x)$ is non-decreasing. Hence, the maximum must occur precisely at $x = 1$. Thus, since you already found that $f(1) = \frac{1}{3}$, the maximum is ⅓.
A: Notice $$f’(x)=x^2 +\int_0^x \frac{\partial}{\partial x}\sqrt{t^4+(1-x)^2(x-t)^2}dt \\ = x^2 +\int_0^x\frac{(1-x)^2\cdot2\cdot(x-t)+(x-t)^2\cdot2\cdot(1-x)}{2\sqrt{t^4+(1-x)^2(x-t)^2}}$$
Now, $x^2\ge 0$, inside the integral the denominator is positive, $2(1-x)^2(x-t)\ge 0$ as $t\le x$. Also, $2(x-t)^2(1-x)\ge 0$ as $x\le 1$. Thus, the integral is positive and $$f’(x)\ge 0$$ in $[0,1]$ $$\implies f_{\text{max}} =f(1) =\frac 13$$
