Question

the integral from -2 to 1 of 1/(x^2 +4x+13)

Answer 1

recall that

∫ du/(a^2+u^2) = 1/a arctan(u/a)

now note that
x^2+4x+13 = (x+2)^2 + 3^2

Nuff said?
I get π/12