Question

Inregrala de la -2 la 0 din 1/(x^2+4x+8)

Answer 1

x²+4x+8 = (x+2)² + 4
(x+a)² +b = x²+2ax+a²+b
2ax = 4x => 2a = 4 => a=2
a²+b = 8 => 4+b = 8 => b = 4
Integrala din $\frac{1}{ x^{2} +4x+8}$ dx = integrala din $\frac{1}{( x+2)^{2} +4}$ dx = $\frac{1}{2} * arctg \frac{x+2}{2}$ luata de la -2 la 0 
=$\frac{1}{2} arctg 1 - \frac{1}{2} arctg 0$