Find the value of $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ Let $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ .
Find the value of $[I+\sqrt{3}]$ (where [.] represents the greatest integer function) 
I have no idea of calculating such type of problem where integration is complex. 

$x^3-3x^2+3x-1\ge x^3-3x²+5 \ge 1$. (for x>2)
$\frac{1}{\sqrt{(x-1)^3}} \le \frac{1}{\sqrt{x^3-3x^+5}} \le 1$ (for x>2)
$\int_2^3 \frac{1}{\sqrt{(x-1)^3}}dx \le \int_2^3\frac{1}{\sqrt{x^3-3x^+5}}dx \le \int_2^3 1dx$
$2-\sqrt{2} \le I \le 1$
$2-\sqrt{2} + \sqrt{3} \le I+\sqrt{3} \le 1+\sqrt{3}$
$2 \lt  I+\sqrt{3} \lt 3$
Hence answer is 2, i don't think the integral have any nice closed form.
