Find the value $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ Let $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ find the value of $\left[ {I + \sqrt 3 } \right] $ {where [.] represent greatest integral function}
Let $T(x)={{x^3} - 3{x^2} + 5}$, $T'\left( x \right) = {x^3} - 3{x^2} + 5 = 3{x^2} - 6x = 3x\left( {x - 2} \right) > 0,x \in \left( {2,3} \right)$
$T(x)$ is increasing for $x\in(2,3)$
Not able to proceed further

Built around $x=\frac 52$ the series expansion of the integrand is
$$2 \sqrt{\frac{2}{15}}-2 \sqrt{\frac{2}{15}} \left(x-\frac{5}{2}\right)+\frac{1}{5}
   \sqrt{\frac{6}{5}} \left(x-\frac{5}{2}\right)^2+\frac{1}{3} \sqrt{\frac{10}{3}}
   \left(x-\frac{5}{2}\right)^3-\frac{111}{50} \sqrt{\frac{3}{10}}
   \left(x-\frac{5}{2}\right)^4+O\left(\left(x-\frac{5}{2}\right)^5\right)$$
Integrate (some terms will disappear because of the symmetry) and you have a very good approximation.
