1. Simplify the floor terms

We need to evaluate $I=\int_4^{10} \frac{[x^2]\,dx}{[x^2-28x+196]+[x^2]}.$

Notice that $x^2-28x+196=(x-14)^2.$ So $I=\int_4^{10} \frac{[x^2]}{[(x-14)^2]+[x^2]}\,dx.$

Also, $(x-14)^2=x^2-28x+196.$ For $x\in[4,10]$, we have $x^2\in[16,100]$ and $(x-14)^2=(14-x)^2\in[16,100].$

2. Key observation

Let $y=x^2.$ Then on the interval $[4,10]$, $x$ is an integer only at isolated points, so floor behavior is easy to inspect through intervals between integers of $x^2$ and $(x-14)^2$.

But there is a much simpler identity: $x^2-28x+196=(14-x)^2.$ Now make the substitution $x=14-t.$ Then as $x$ goes from $4$ to $10$, $t$ goes from $10$ to $4$.

Define $f(x)=\frac{[x^2]}{[x^2]+[(14-x)^2]}.$ Then $f(14-x)=\frac{[(14-x)^2]}{[(14-x)^2]+[x^2]}.$ Hence, $f(x)+f(14-x)=1$ for all $x$ where defined, and here it is defined everywhere on $[4,10]$ because denominator is positive.

3. Use symmetry of the interval

The interval $[4,10]$ is symmetric about $x=7$, and the transformation $x\mapsto 14-x$ maps $[4,10]$ onto itself.

Therefore, $I=\int_4^{10} f(x)\,dx.$ Also, $I=\int_4^{10} f(14-x)\,dx.$ Adding,