Question

Resolva a seguinte integral:

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

Answer 1

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.

Answer 2

Integrais que produzem funções trigonométricas inversas

$\boxed{\boxed{\boxed{\mathsf{\displaystyle\int\frac{du}{{a}^{2}+{u}^{2}}} = \dfrac{1}{a}arctg( \frac{u}{a}) + c}}}$

$\displaystyle\mathsf{\int\dfrac{dx}{3+{x}^{2}}=\dfrac{1}{\sqrt{3}}arctg(\dfrac{x}{\sqrt{3}})+c}$