Resolva a seguinte integral:.... Pergunta de ideia deKrikor

November 2019 1 163 Report

Resolva a seguinte integral:

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

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Olá Krikor.

Identidade utilizada:

$\star~~\boxed{\boxed{\mathsf{\dfrac{d}{dx}~b\cdot arctan(ax)=\dfrac{ba}{(ax)^2+1}}}}$

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Organizando e resolvendo a integral

$\mathsf{\displaystyle\int \dfrac{1}{x^2+3}~dx=\int\dfrac{1}{3\cdot\Big(\dfrac{x^2}{3}+1\Big)}~dx=\mathsf{\displaystyle\dfrac{1}{3}\cdot\int\dfrac{1}{\dfrac{x^2}{3}+1}}~dx}}$

Pela identidade acima fazendo $\mathsf{a^2=\frac{1}{3}}$ e $\mathsf{b=1}$ temos:

$\mathsf{\dfrac{d}{dx}~arctan(\dfrac{x}{\sqrt{3}})=\dfrac{\dfrac{1}{\sqrt{3}}}{\dfrac{x}{3}+1}}\\\\\\\\\mathsf{\dfrac{d}{dx}~\sqrt{3}\cdot arctan(\dfrac{x}{\sqrt{3}})=\dfrac{1}{\dfrac{x^2}{3}+1}}}$

Integrando em ambos os lados, temos:

$\mathsf{\displaystyle\int\dfrac{1}{\dfrac{x^2}{3}+1}~dx=\displaystyle\int\dfrac{d}{dx}~\sqrt{3}\cdot arctan(\dfrac{x}{\sqrt{3}})~dx}\\\\\\\\\mathsf{\displaystyle\int\dfrac{1}{\dfrac{x^2}{3}+1}~dx =\sqrt{3}\cdot arctan(\dfrac{x}{\sqrt{3}})}$

Multiplique ambos os lados por $\frac{1}{3}$

$\mathsf{\displaystyle \dfrac{1}{3}\cdot\int \dfrac{1}{\dfrac{x^2}{3}+1}~dx=\dfrac{\sqrt{3}}{3}\cdot arctan(\dfrac{x}{\sqrt{3}})}\\\\\\\mathsf{\displaystyle\int\dfrac{1}{x^2+3}~dx=\dfrac{\sqrt{3}}{3}\cdot arctan(\dfrac{x}{\sqrt{3}})+C}$

Dúvidas? comente.

Krikor Muito obrigado! :)

superaks Disponha !

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