Question

Resolva a seguinte integral:

$\displaystyle\int \dfrac{-2x^3 + 4}{x(x^2 + 3) }$

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Answer 1

Resposta:

$\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4\ln|x|}{3}-\dfrac{2\ln|x^2+3|}{3}+2\sqrt{3}\arctan\left(\dfrac{x}{\sqrt{3}}\right)\right)+C$

Resolução:

Podemos resolver este exercício usando as Regras de Integração, Simplificando algumas Frações e usando Substituições.

    $\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2\displaystyle\int{\dfrac{x^3-2}{x(x^2+3)}\;dx}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2\displaystyle\int{\left(1+\dfrac{-3x-2}{x(x^2+3)}}\;dx\right)}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2\left(\displaystyle\int{1\;\;dx}+\displaystyle\int{\left(\dfrac{-3x-2}{x(x^2+3)}}\;dx\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x-2\displaystyle\int{\left(\dfrac{-3x-2}{x(x^2+3)}}\;dx\right)}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x-2\left(\displaystyle\int{\left(-\dfrac{2}{3x}+\dfrac{2x-9}{3(x^2+3)}\;dx\right)\right)}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x-2\left(\displaystyle\int{\left(-\dfrac{2}{3x}\;dx\right)}+\displaystyle\int{\left(\dfrac{2x-9}{3(x^2+3)}\;dx\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\displaystyle\int{\left(\dfrac{1}{x}\;dx\right)}-\dfrac{2}{3}\displaystyle\int{\left(\dfrac{2x-9}{x^2+3}\;dx\right)}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\displaystyle\int{\left(\dfrac{2x}{x^2+3}-\dfrac{9}{x^2+3}\;dx\right)}\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\displaystyle\int{\left(\dfrac{2x}{x^2+3}\;dx\right)}-\displaystyle\int{\left(\dfrac{9}{x^2+3}\;dx\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln{(x^2+3)}-9\displaystyle\int{\left(\dfrac{1}{x^2+3}\;dx\right)}\right)$

Tenha-se:

$u=\dfrac{x}{\sqrt{3}}\;\longrightarrow\;\dfrac{du}{dx}=\dfrac{1}{\sqrt{3}}\Leftrightarrow dx=\sqrt{3}\;du$

Logo,

    $\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-9\displaystyle\int{\left(\dfrac{1}{x^2+3}\;dx\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-9\displaystyle\int{\left(\dfrac{1}{3u^2+3}\sqrt{3}\;du\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-9\displaystyle\int{\left(\dfrac{\sqrt{3}}{3u^2+3}\;du\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-9\times\dfrac{\sqrt{3}}{3}\displaystyle\int{\left(\dfrac{1}{u^2+1}\;du\right)}\right)\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-3\sqrt{3}\arctan{u}\right)+C\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4}{3}\ln|x|-\dfrac{2}{3}\left(\ln|x^2+3|-3\sqrt{3}\arctan\left(\dfrac{x}{\sqrt{3}} \right)\right)+C\Leftrightarrow$

$\Leftrightarrow\displaystyle\int{\dfrac{-2x^3+4}{x(x^2+3)}\;dx}=-2x+\dfrac{4\ln|x|}{3}-\dfrac{2\ln|x^2+3|}{3}+2\sqrt{3}\arctan\left(\dfrac{x}{\sqrt{3}}\right)\right)+C$

Por falta de espaço, podes encontrar os cálculos auxiliares no 1º anexo desta resposta.

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