I=∫2xdx(x^4-1) x∈[√2;√3] face substitutia x²=y 2xdx=dy
se recvalculeaza limitele de integrare pt x=√2 =>y=2 x=√3 y=3
integrala devine
I=∫dy/(y²-1)=∫dy/(y+1)(y-1) unde y∈[2,3]
1/(y-1)(y+1)=A/(y-1)+B/(y+1)=(Ay+A+BY-B)/(y-1)(y+1)=[(A+B)y+(A-B)]/))Y-1)(y+1)
Folosim metoda identificarii
{A+B=0
{A-B=1 =>A=1/2 B=- 1/2
Integrala devine
I=1/2∫dy/(y+1)-1/2∫dy/(y-1)=1/2[ln(y+1)-ln(y-1)]=1/2ln(y+1)/(y-1) y∈[2.3]
I=ln(3+1)/(3-1)-ln(2+1)/(2-1)=ln2-ln3=ln2/3
