Metode calcul integrale definite Integrarea functiilor rationale

Inainte de a vorbi despre Integrarea functiilor rationale o sa aflam metode de calcul a integralelor definite

Dupa ce am invatat sa rezolvam integralele prin metoda integrarii prin parti, dar si prin metoda schimbarii de variabile, a venit vremea sa discutam despre Integrarea functiilor rationale, dar si Metode de calcul a integralelor definite.

Definitie: O functie rationala f definita pe un interval I este de forma $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}, \forall x\in I$ , unde $P, R\in R, Q\left(x\right)\neq 0$ pe I.
O functie rationala se numeste functie rationala simpla, daca are una din formele:
– $f\left(x\right)=P\left(x\right), P\in R\left[X\right]$
– $f\left(x\right)=\frac{1}{\left(x-a\right)^{n}}, A, a\in R, n\in N^{*}$
– $f\left(x\right)=\frac{Ax+B}{\left(x^{2}+ax+b\right)^{n}}, A, b, a, b\in R, \Delta =b^{2}-4\cdot a\cdot c<0, n\in N^{*}$
Teorema: Orice functie rationala se poate descompune in mod unic, in suma de functii rationale simple.
Exercitii cu integrarea functiilor rationale :
1) Sa se calculeze:

a) $\int\frac{x+1}{x^{3}-4x^{2}+5x-2}dx$
Avem
$Q\left(x\right)=x^{3}-4x^{2}+5x-2$
Ca sa descompunem polinomul Q in produs de factori primi, cautam printre divizorii permenului liber. Astfel avem
$D_{-2}=\left\{\pm 1; \pm 2\right\}$
Incepem sa verificam pentru: 1
$Q\left(1\right)=1^{3}-4\cdot 1^{2}+5\cdot 1-2=1-4+5-2=-3+3=0$
Deci 1 este radacina a polinomului
Dar polinomul de mai sus mai putel sa- l scriem si astfel
$Q\left(x\right)=x^{3}-4x^{2}+5x-2=x^{3}-2x^{2}-2x^{2}+4x+x-2=x^{2}\left(x-2\right)-2x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^{2}-2x+1\right)=\left(x-2\right)\left(x-1\right)^{2}$ .
Astfel integrala putem sa o scriem:
$\int\frac{x+1}{\left(x-2\right)\left(x-1\right)^{2}}dx=\int\left(\frac{A}{x-2}+\frac{B}{x-1}+\frac{C}{\left(x-1\right)^{2}}\right)dx=\int\frac{A\left(x-1\right)^{2}+B\left(x-1\right)\left(x-2\right)+C\left(x-2\right)}{\left(x-2\right)\left(x-1\right)^{2}}dx$
Dupa ce am adus la acelasi numitor gasim
$x+1=A\left(x-1\right)^{2}+B\left(x-1\right)\left(x-2\right)+C\left(x-2\right) \\ x+1=A\left(x^{2}-2x+1\right)+B\left(x^{2}-2x-x+2\right)+Cx-2C \\x+1=Ax^{2}-2Ax+A+Bx^{2}-3Bx+2B+Cx-2C \\x+1=x^{2}\left(A+B\right)+x\left(-2A-3B+C\right)+A+2B-2C$

Astfel gasim sistemul
$A+B=0 \\-2A-3B+C=1 \\A+2B-2C=1 \\A=-B \\ -2\cdot \left(-B\right)-3B+C=1\Rightarrow 2B-3B+C=1\Rightarrow -B+C=1\Rightarrow C=1+B \\-B+2B-2C=1\Rightarrow B-2C=1\Rightarrow B-2\left(1+B\right)=1\Rightarrow B-2-2B=1\Rightarrow -B=1+2\Rightarrow B=-3$
Mai stim si ca $A=-B\Rightarrow A=-\left(-3\right)\Rightarrow A=3$
Acum ca sa aflam C stim ca $C=1+B\Rightarrow C=1+\left(-3\right)\Rightarrow C=1-3\Rightarrow C=-2$
Astfel obtinem integralele
$\int\frac{3}{x-2}dx+\int\frac{-3}{x-1}dx+\int\frac{-2}{\left(x-1\right)^{2}}dx=3\ln\left(x-2\right)-3\ln\left(x-1\right)-\frac{-2}{\left(2-1\right)\left(x-1\right)^{2-1}}=3\ln\left(x-2\right)-3\ln\left(x-1\right)+\frac{2}{x-1}+C$
Sa rezolvam acum si o integrala definita
$\int^{2}_{0}\frac{x^{2}-1}{x^{3}+5x^{2}+8x+4}dx$
Incercam sa descompunem functiile rationale simple:
Astfel avem
$x^{3}+5x^{2}+8x+4=x^{3}+x^{2}+4x^{2}+4x+4x+4=x^{2}\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x^{2}+4x+4\right)=\left(x+1\right)\left(x+2\right)^{2}$ .
Astfel integrala devine
$\int^{2}_{0}\frac{x^{2}-1}{x^{3}+5x^{2}+8x+4}dx=\int^{2}_{0}\left(\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{\left(x+2\right)^{2}}\right)dx=\int^{2}_{0}\left(\frac{A\left(x+2\right)^{2}+B\left(x+1\right)\left(x+2\right)+C\left(x+1\right)}{\left(x+1\right)\left(x+2\right)^{2}}\right)dx= \int^{2}_{0}\left(\frac{A\left(x^{2}+4x+4\right)+B\left(x^{2}+2x+x+2\right)+Cx+C}{\left(x+1\right)\left(x+2\right)^{2}}\right)dx= \int^{2}_{0}\left(\frac{Ax^{2}+4Ax+4A+Bx^{2}+3Bx+2B+Cx+C}{\left(x+1\right)\left(x+2\right)^{2}}\right)dx= \int^{2}_{0}\left(\frac{x^{2}\left(A+B\right)+x\left(4A+3B+C\right)+4A+2B+C}{\left(x+1\right)\left(x+2\right)^{2}}\right)dx$
Astfel obtinem sistemul de ecuatii
$A+B=1 \\4A+3B+C=0 \\4A+2B+C=-1$
Acum rezolvam sistemul, din prima ecuatie scoatem A in functie de B si inlocuim in urmatoarele ecuatii
$A=1-B$
$4\left(1-B\right)+3B+C=0\Rightarrow 4-4B+3B+C=0\Rightarrow -B+C=-4\Rightarrow C=-4+B\Rightarrow C=B-4 \\4\left(1-B\right)+2B+C=-1\Rightarrow 4-4B+2B+C=1\Rightarrow 4-2B+C=-1\Rightarrow -2B+C=-1-4\Rightarrow -2B+C=-5\Rightarrow -2B+B-4=-3\Rightarrow -B=-3+4\Rightarrow -B=1\Rightarrow B=-1$
Acum daca stim B putem afla C
$C=-1-4\Rightarrow C=-5$
Acum putem afla A, deoarece stim ca
$A+B=1\Rightarrow A+\left(-1\right)=1\Rightarrow A=1+1\Rightarrow A=2$
Astfel integrala devine:
$\int^{2}_{0}\frac{2}{x+1}dx+\int^{2}_{0}\frac{-1}{x+2}dx+\int^{2}_{0}\frac{-5}{\left(x+2\right)^{2}}=$

$2\ln\left(x+1\right)|^{2}_{0}-\ln\left(x+2\right)|^{2}_{0}-\frac{-5}{\left(2-1\right)\left(x+2\right)^{2-1}}|^{2}_{0}=$

$2\ln 3-2\ln 1-\left(\ln 4-\ln 2\right)+\frac{5}{1\cdot\left(x+2\right)}|^{2}_{0}=2\ln 3-\ln 2+\frac{5}{4}-\frac{5}{2}=$

$2\ln 3-\ln 2+\frac{5-2\cdot 5}{4}=2\ln 3-\ln 2-\frac{5}{4}=\ln 3^{2}-\ln 2-\frac{5}{4}=\ln\frac{9}{2}-\frac{5}{4}$ .
c) $\int^{3}_{2}\frac{x^{2}}{x^{2}-1}dx=\int^{3}_{2}\frac{x^{2}-1+1}{x^{2}-1}dx=$
$\int^{3}_{2}\frac{x^{2}-1}{x^{2}-1}dx+\int^{3}_{2}\frac{1}{x^{2}-1}dx=$
$\int^{3}_{2}dx+\frac{1}{2\cdot 1}\ln|\frac{x-1}{x+1}||^{3}_{2}=$
$x|^{3}_{2}+\frac{1}{2\cdot 1}\ln|\frac{x-1}{x+1}||^{3}_{2}=$
$3-2+\frac{1}{2}\left(\ln|\frac{3-1}{3+1}|-\ln|\frac{2-1}{2+1}|\right)=$
$1+\frac{1}{2}\left(\ln|\frac{2}{4}|-\ln|\frac{1}{3}|\right)=$
$1+\frac{1}{2}\ln|\frac{1}{2}\cdot \frac{3}{1}|=1+\frac{1}{2}\ln\frac{3}{2}$ .
Ca sa rezolvam integrala de mai sus am scazut la numarator cifra 1 si am adunat, apoi am impartit inegrala in doua parti, prima observam ca am simplificat-o iar cea de-a doua am rezolvat-o cu ajutorul formulelor din tabel pentru integrare, iar apoi am aplicat Leibniz-Newton.