a)observi ca (x²+3x+3)*(2x+3)/(x²+3x+3)=2x+3
x∈[1,2] I=∫(2x+3)dx=∫2xdx+∫3dx=(x²+3x)/1↑2=(4+6)-(1+3)=10-4=6
b)Aria=∫(2x+3)dx/(x²+3x+3)
x²+3x+3=y =>(2x+3)dx/(x²+3x+3)=dy
x=0 y=3 x=3 y=21
Aria=∫dy/y= y∈[3,21]
Aria=lny/3↑21=ln21-ln3=ln21/3=ln7
c)Integrala se rezolva prin parti x∈[-1,0]
u=f(x) du=f `(x)dx
dv=f `(x)dx=> v=f(x)
I=f(x)*f(x)-∫f(x)*f`(x)dx
I=f²(x)-I
2I=f²(x)
I=f²(x)/2
I=(2x+3)²/2*(x²+3x+3)²/(-1)↑0=9/18-(-1)²/2*(-1)=1/2-1/2=0
