Integral $\int_{-3}^2 \frac{[x]}{[x]^2+2[x]+x^3}dx$ I came across this question which was supposed to be solved in about $1$ or $2$ minutes, but I came across a severe roadblock.
The question was:

Integrate $$\int_{-3}^2 \frac{\lfloor x\rfloor}{\lfloor x\rfloor^2+2\lfloor x \rfloor+x^3}dx$$ 

I split the integral into $5$ separate ones to substitute a known value of step $x$. But following that I had to use partial fractions with some weird numbers and that made it pretty long. How do I approach this sum? 

Yes you are right we have to split the integral into 5:
$$I_1 = \int_{-3}^{-2}\frac{-3dx}{3 + x^3}$$
$$I_2 = \int_{-2}^{-1}\frac{-2dx}{x^3} = \frac{3}{4}$$
$$I_3 = \int_{-1}^{0}\frac{-dx}{-1 + x^3}$$
$$I_4 = 0$$
$$I_5 = \int_{1}^{2}\frac{dx}{3 + x^3}$$
Separately we try to find the integral 
$$\int\frac{dx}{a^3 + x^3}$$
$$= \int(\frac{1}{3a^2(x + a)} + \frac{1}{3a^2}\frac{- x + 2a}{x^2 -ax + a^2})dx$$
$$= \int\frac{1}{3a^2}(\frac{1}{x + a} - \frac{x - \frac{a}{2} - \frac{3a}{2}}{(x - \frac{a}{2})^2 + (\frac{\sqrt{3}a}{2})^2})dx$$
$$= \frac{1}{6a^2}ln\frac{(x + a)^2}{x^2 -ax + a^2} + \frac{1}{\sqrt{3}a^2}tan^{-1}\frac{2x - a}{\sqrt{3}a}$$
Using this formula we have 
$$I = I_1 + I_2 + I_3 + I_4 + I_5 = 0.2748 + 0.75 + 0.8356 + 0 + 0.1615 = 2.0219$$
