descompunem numitorul in factori
x⁴-x²= x²(x-1)(x+1)
folosim metoda coeficientilor nedeterminati
(3x³-2x²-x+2)/ x²(x-1)(x+1)= A/x+ B/x²+ C/(x-1)+ D/(x+1)
aducem la acelasi numitor si, facand inmultirile, obtinem
(Ax³-Ax+Bx²-B+Cx³+Cx²+Dx³-Dx²)/ x²(x-1)(x+1)
dam factori puterile lui x
[x³(A+C+D)+ x²(B+C-D)+ X(-A)-B]/ x²(x-1)(x+1)
identificam coeficientii
A+C+D= 3
B+C- D= -2
-A= -1
-B= 2
de unde obtinem
A=1, B= -2, C= D= 1
inlocuim in descompunerea initiala si obtinem
integrala= ∫1/x dx - 2 ∫1/x² dx+ ∫1/(x-1) dx+ ∫1/x² dx=
= ln|x|+ 2/x+ ln|x-1|+ ln|x+1|+ C=
= 2/x + ln|x(x-1)(x+1)|+ C
mai verifica o data calculele
O seara buna!
