<s>[INST] Integrate $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = $ Solve $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx}  = \_\_\_\_\_$ where [.] represent greatest integer function.
My approch is as follow
$\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{\left[ {{{\left( {x - 4} \right)}^2}} \right] + \left[ {{x^2}} \right]}}dx} $
$y=(x-4)^2$ represent a quadratic equation
Refer to the image below

We see that for some x value y will have values of 9,8,7,....,1
SO how we will proceed this
I presume it will be a telescopic function
 [/INST] $$I=\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {(x-4)^2} \right]}} + \left[ {{x^2}} \right]}}dx}!$$
Use $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$$, then
$$I=\int\limits_1^3 {\frac{{\left[ {{(x-4)^2}} \right]}}{{{{\left[ {x^2} \right]}} + \left[ {{(x-4)^2}} \right]}}dx}$$
Adding the two we get $$2I=\int_{1}^{3} dx \implies I=1.$$
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