Question

Sa se calculeze integrala urmatoare :

Answer 1

x^3+X^2-2=x^3-x^2+2x^2-2=x^2(x-1)+2(x+1)(x-1)=(x-1)(x^2+2x+2)
Se aplica teoria de la integrarea fractiilor cu polinoame
=(3x^2-5x+3)/(x-1)(x^2+2x+2)=A/(x-1)+(Bx+C)/(x^2+2x+2)=(Ax^2+2Ax+2A+Bx^2-Bx+Cx-C)/(x-1)(x^2+2x+2)
Grupam termenii si facem egalitatile
A+B=3
2A-B+C=-5
2A-C=3  care furnizeaza solutiile
A=1/5   B=14/5   C=-13/5
(3x^2-5x+3)/(x^3+X^2-2)=1/5*1/(x-1)+(14x/5-13/5)/(x^2+2x+2)
integrala din enunt s-a impartit in alte 2
I1=1/5*S[1/(x-1)]dx=1/5*lnIx-1I+C
I2=1/5*S[(14x-13)/(x^2+2x+2)]dx=1/5*S[(14x+14-27)/(x^2+2x+2)]dx=1/5*{7*S[(2x+2)/(x^2+2x+2)]dx-S[27/(x^2+2x+2)]dx} = 7/5*ln(x^2+2x+2)-27/5*S[1/[(x+1)^2+1]dx
integrala S[1/[(x+1)^2+1]dx=arctg(x+1)+C2
Final   I=1/5*lnIx-1I+7/5*ln(x^2+2x+2)-27/5*arctg(x+1)+C