Question

resolva a integral pelo método de frações parciais:
integral de x^3 + 3x - 1/x^4 - 4x^2

Answer 1

$\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}=\dfrac{x^{3}+3x-1}{x^{2}(x^{2}-4)}=\dfrac{x^{3}+3x-1}{x^{2}(x^{2}-2^{2})}=\dfrac{x^{3}+3x-1}{x^{2}(x+2)(x-2)}$

Devemos escrever a fração acima da seguinte forma:

$\dfrac{x^{3}+3x-1}{x^{2}(x+2)(x-2)}=\dfrac{A}{x}+\dfrac{B}{x^{2}}+\dfrac{C}{x+2}+\dfrac{D}{x-2}$

Manipulando a parte da direita:

$\dfrac{Ax(x+2)(x-2)+B(x+2)(x-2)+Cx^{2}(x-2)+Dx^{2}(x+2)}{x^{2}(x+2)(x-2)}\\\\\\\dfrac{Ax(x^{2}-4)+B(x^{2}-4)+C(x^{3}-2x^{2})+D(x^{3}+2x^{2})}{x^{2}(x+2)(x-2)}\\\\\\\dfrac{Ax^{3}-4Ax+Bx^{2}-4B+Cx^{3}-2Cx^{2}+Dx^{3}+2Dx^{2}}{x^{2}(x+2)(x-2)}\\\\\\\dfrac{(A+C+D)x^{3}+(B-2C+2D)x^{2}-4Ax-4B}{x^{2}(x+2)(x-2)}$
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Então:

$\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}=\dfrac{(A+C+D)x^{3}+(B-2C+2D)x^{2}-4Ax-4B}{x^{2}(x+2)(x-2)}$

Cortando os denominadores (já que são iguais), temos:

$x^{3}+3x-1=(A+C+D)x^{3}+(B-2C+2D)x^{2}-4Ax-4B\\1x^{3}+0x^{2}+3x-1=(A+C+D)x^{3}+(B-2C+2D)x^{2}-4Ax-4B$

Os polinômios só serão iguais se:

$\begin{cases}A+C+D=1\\B-2C+2D=0\\-4A=3\\-4B=-1\end{cases}$

Achando A e B:

$-4A=3~~~\therefore~~~4A=-3~~~\therefore~~~\boxed{\boxed{A=-\dfrac{3}{4}}}\\\\\\-4B=-1~~~\therefore~~~4B=1~~~\therefore~~~\boxed{\boxed{B=\dfrac{1}{4}}}$

Então:

$A+C+D=1~~~\therefore~~~-\dfrac{3}{4}+C+D=1~~~\therefore~~~C+D=\dfrac{7}{4}\\\\B-2C+2D=0~~~\therefore~~~\dfrac{1}{4}-2C+2D=0~~~\therefore~~~-C+D=-\dfrac{1}{8}$

Então:

$\begin{cases}C+D=\dfrac{7}{4}\\-C+D=-\dfrac{1}{8}\end{cases}$

Somando as equações:

$2D=\dfrac{7}{4}-\dfrac{1}{8}=\dfrac{14}{8}-\dfrac{1}{8}=\dfrac{13}{8}$

Logo:

$\boxed{\boxed{D=\dfrac{13}{16}}}$

Achando C:

$C=\dfrac{7}{4}-\dfrac{13}{16}=\dfrac{28}{16}-\dfrac{13}{16}=\dfrac{15}{16}$

Portanto:

$\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}=\dfrac{(-\frac{3}{4})}{x}+\dfrac{(\frac{1}{4})}{x^{2}}+\dfrac{(\frac{13}{16})}{x+2}+\dfrac{(\frac{13}{16})}{x-2}$
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Finalmente:

$\int\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}dx=\int\dfrac{(-\frac{3}{4})}{x}+\dfrac{(\frac{1}{4})}{x^{2}}+\dfrac{(\frac{13}{16})}{x+2}+\dfrac{(\frac{13}{16})}{x-2}dx\\\\\\\int\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}dx=\int\dfrac{(-\frac{3}{4})}{x}dx+\int\dfrac{(\frac{1}{4})}{x^{2}}dx+\int\dfrac{(\frac{15}{16})}{x+2}dx+\int\dfrac{(\frac{13}{16})}{x-2}dx$

Essas integrais são resolvidas por substituição de variáveis (com exceção das duas primeiras):

$\int\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}dx=-\dfrac{3}{4}ln|x|+\int\dfrac{(\frac{1}{4})}{x^{2}}dx+\dfrac{15}{16}ln|x+2|+\dfrac{13}{16}ln|x-2|+C_{1}$

A integral que sobrou é resolvida normalmente:

$\int\dfrac{(\frac{1}{4})}{x^{2}}dx=\dfrac{1}{4}\int x^{-2}dx=\dfrac{1}{4}\dfrac{x^{-2+1}}{(-2+1)}=-\dfrac{1}{4x}$

Portanto:

$\boxed{\boxed{\int\dfrac{x^{3}+3x-1}{x^{4}-4x^{2}}dx=-\dfrac{3}{4}ln|x|-\dfrac{1}{4x}+\dfrac{15}{16}ln|x+2|+\dfrac{13}{16}ln|x-2|+C}}$