Question

Determine a integral indefinida de (x^6- x^3 + 1)/(x^4 - 9x^2) dx

Answer 1

Resposta:

Explicação passo-a-passo:

Questão Problema

${\displaystyle\int}\dfrac{x^6-x^3+1}{x^4-9x^2}\,\mathrm{d}x$

dividindo o polinomio

$={\displaystyle\int}\left(\dfrac{-x^3+81x^2+1}{x^4-9x^2}+x^2+9\right)\mathrm{d}x$

Aplicando a lineariedade

$=-{\displaystyle\int}\dfrac{x^3-81x^2-1}{x^4-9x^2}\,\mathrm{d}x+{\displaystyle\int}x^2\,\mathrm{d}x+{9}}{\displaystyle\int}1\,\mathrm{d}x$

Resolvendo${\displaystyle\int}\dfrac{x^3-81x^2-1}{x^4-9x^2}\,\mathrm{d}x\\fatorando\\={\displaystyle\int}\dfrac{x^3-81x^2-1}{\left(x-3\right)x^2\left(x+3\right)}\,\mathrm{d}x$

Realizando a decomposição em parciais$={\displaystyle\int}\left(\dfrac{757}{54\left(x+3\right)}+\dfrac{1}{9x^2}-\dfrac{703}{54\left(x-3\right)}\right)\mathrm{d}x$

Aplicando a linearidade

$=\dfrac{757}{54}}}{\displaystyle\int}\dfrac{1}{x+3}\,\mathrm{d}x+{\dfrac{1}{9}}}{\displaystyle\int}\dfrac{1}{x^2}\,\mathrm{d}x-}{\dfrac{703}{54}}}{\displaystyle\int}\dfrac{1}{x-3}\,\mathrm{d}x\\resolvendo\\{\displaystyle\int}\dfrac{1}{x+3}\,\mathrm{d}x\\substituindo\rightarrow\,u=x+3\rightarrow\,\,\frac{du}{dx}\,\,\rightarrow\,\,du=dx\\={\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u\\=\ln\left(u\right)\\=\ln\left(x+3\right)$

terminando os calculos

chegará nesta resposta

$-\dfrac{757\ln\left(\left|x+3\right|\right)-18x^3-486x-\frac{6}{x}-703\ln\left(\left|x-3\right|\right)}{54}$