The value of the integral $$\int_4^{10} \frac{[x^2]}{[x^2 - 28x + 196] + [x^2]}dx$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is

We have to evaluate the definite integral

$$I=\int_{4}^{10}\dfrac{\left\lfloor x^{2}\right\rfloor}{\left\lfloor x^{2}-28x+196\right\rfloor+\left\lfloor x^{2}\right\rfloor}\,dx$$

First note the algebraic identity

$$x^{2}-28x+196=(x-14)^{2}\;,$$

so the denominator contains the term $$\left\lfloor(x-14)^{2}\right\rfloor$$. Define a new function

$$f(x)=\dfrac{\left\lfloor x^{2}\right\rfloor}{\left\lfloor x^{2}\right\rfloor+\left\lfloor(14-x)^{2}\right\rfloor}\;.$$ With this notation the integral becomes simply

$$I=\int_{4}^{10}f(x)\,dx\;.$$

Now observe a fundamental symmetry. For every real number $$x$$ in the interval $$[4,10]$$, the point $$14-x$$ also lies in the same interval, because

$$4\le x\le 10\quad\Longrightarrow\quad 4\le 14-x\le 10.$$

Therefore we can look at the value of the integrand at the “mirror” point $$14-x$$. We write

$$f(14-x)=\dfrac{\left\lfloor(14-x)^{2}\right\rfloor}{\left\lfloor(14-x)^{2}\right\rfloor+\left\lfloor x^{2}\right\rfloor}\;.$$

Add the two expressions:

$$f(x)+f(14-x)= \dfrac{\left\lfloor x^{2}\right\rfloor}{\left\lfloor x^{2}\right\rfloor+\left\lfloor(14-x)^{2}\right\rfloor} +\dfrac{\left\lfloor(14-x)^{2}\right\rfloor}{\left\lfloor x^{2}\right\rfloor+\left\lfloor(14-x)^{2}\right\rfloor}.$$

Because the two fractions share the same denominator, we can combine their numerators directly:

$$f(x)+f(14-x)= \dfrac{\left\lfloor x^{2}\right\rfloor+\left\lfloor(14-x)^{2}\right\rfloor} {\left\lfloor x^{2}\right\rfloor+\left\lfloor(14-x)^{2}\right\rfloor}=1.$$

So for every $$x$$ in $$[4,10]$$ we have the simple relation

$$f(x)+f(14-x)=1.$$

Next, evaluate the integral $$I$$ once more, but this time make the substitution

$$u=14-x\quad\Longrightarrow\quad du=-dx.$$

When $$x=4$$, $$u=10$$, and when $$x=10$$, $$u=4$$. Hence

$$I=\int_{4}^{10}f(x)\,dx =\int_{4}^{10}f(14-u)\,du =\int_{4}^{10}f(14-u)\,du.$$

Because the dummy variable of integration can be renamed back to $$x$$, this shows

$$I=\int_{4}^{10}f(14-x)\,dx.$$

Add this last equality to the original definition of $$I$$:

$$2I =\int_{4}^{10}\bigl[f(x)+f(14-x)\bigr]\,dx.$$

We have already proved that the integrand inside the brackets equals $$1$$, so the integral simplifies immediately:

$$2I=\int_{4}^{10}1\,dx =(10-4)=6.$$

Dividing by $$2$$ we obtain

$$I=3.$$

Hence, the correct answer is Option D.