Question

Calcule a integral indefinida ∫ x-3 / x2-3x+2.dx Usando o metodo das fraçoes parciais.

Answer 1

$\displaystyle\int\dfrac{x-3}{x^{2}-3x+2}dx$

Calculando as raízes de x² - 3x + 2, achamos x' = 1 e x'' = 2. Portanto:

$x^{2}-3x+2=(x-x')\cdot(x-x'')=(x-1)\cdot(x-2)$

Então, queremos achar A e B tais que:

$\dfrac{x-3}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}\\\\\\\dfrac{x-3}{(x-1)(x-2)}=\dfrac{A(x-2)+B(x-1)}{(x-1)(x-2)}\\\\\\\dfrac{x-3}{(x-1)(x-2)}=\dfrac{Ax-2A+Bx-B}{(x-1)(x-2)}\\\\\\\dfrac{x-3}{(x-1)(x-2)}=\dfrac{(A+B)x-(2A+B)}{(x-1)(x-2)}$

Cancelando os denominadores (assumindo x ≠ 1 e x ≠ 2):

$(A+B)X-(2A+B)=x-3$

Dois polinômios são iguais se e somente se os coeficientes de cada termo são iguais. Portanto:

$A+B=1\\\\-(2A+B)=-3~~~\therefore~~~2A+B=3$

Temos o seguinte sistema para resolver:

$\begin{cases}A+B=1\\2A+B=3\end{cases}$

Subtraindo membro a membro:

$2A-A+B-B=3-1\\\\\boxed{\boxed{A=2}}$

Achando B:

$A+B=1\\\\2+B=1\\\\\boxed{\boxed{B=-1}}$

Então:

$\boxed{\boxed{\dfrac{x-3}{x^{2}-3x+2}=\dfrac{2}{x-1}-\dfrac{1}{x-2}}}$
_____________________________

$\displaystyle\int\dfrac{x-3}{x^{2}-3x+2}dx=\int\left(\dfrac{2}{x-1}-\dfrac{1}{x-2}\right)dx\\\\\\\int\dfrac{x-3}{x^{2}-3x+2}dx=\int\dfrac{2}{x-1}dx-\int\dfrac{1}{x-2}dx\\\\\\\int\dfrac{x-3}{x^{2}-3x+2}dx=2\int\dfrac{1}{x-1}dx-\int\dfrac{1}{x-2}dx$

Resolvemos essas integrais facilmente: Basta fazermos mudança de variáveis

$u=x-1~~\rightarrow~~du=dx\\\\a=x-2~~\rightarrow~~da=dx$

Logo:

$\displaystyle\int\dfrac{x-3}{x^{2}-3x+2}dx=2\int\dfrac{1}{u}du-\int\dfrac{1}{a}da\\\\\\\int\dfrac{x-3}{x^{2}-3x+2}dx=2\cdot ln|u|-ln|a|+C\\\\\\\boxed{\boxed{\int\dfrac{x-3}{x^{2}-3x+2}dx=2\cdot ln|x-1|-ln|x-2|+C}}$