Question

Por favor me ajudem com essa integral:

$\int{\frac{dx}{2x^{2}+5 } }$

Answer 1

Resposta:  $\mathsf{\dfrac{\sqrt{10}}{10}\,arctg\!\left(\dfrac{2x}{\sqrt{10}}\right)+C.}$

Explicação passo-a-passo:

Calcular a integral indefinida:

    $\mathsf{\displaystyle\int\frac{1}{2x^2+5}\,dx}$

Multiplique o numerador e o denominador por 2:

    $\mathsf{=\displaystyle\int\frac{2}{2\cdot (2x^2+5)}\,dx}\\\\\\ \mathsf{=\displaystyle\int\frac{2}{4x^2+10}\,dx}\\\\\\ \mathsf{=\displaystyle \int\frac{1}{(2x)^2+(\sqrt{10})^2}\cdot 2\,dx}$

Faça a seguinte substituição trigonométrica:

    $\mathsf{2x=\sqrt{10}\,tg\,\theta\quad\Longrightarrow\quad 2\,dx=\sqrt{10}\,sec^2\,\theta\,d\theta}\\\\\\ \mathsf{\Longleftrightarrow\quad tg\,\theta=\dfrac{2x}{\sqrt{10}}}\\\\\\ \mathsf{\Longrightarrow\quad \theta=arctg\!\left(\dfrac{2x}{\sqrt{10}}\right),~~com~-\,\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}}$

Além disso, temos

    $\mathsf{(2x)^2+(\sqrt{10})^2=(\sqrt{10}\,tg\,\theta)^2+(\sqrt{10})^2}\\\\ \mathsf{\Longleftrightarrow\quad (2x)^2+(\sqrt{10})^2=10\,tg^2\,\theta+10}\\\\ \mathsf{\Longleftrightarrow\quad (2x)^2+(\sqrt{10})^2=10\,(tg^2+1)}\\\\ \mathsf{\Longleftrightarrow\quad (2x)^2+(\sqrt{10})^2=10\,sec^2\,\theta\qquad\checkmark}$

Substituindo, a integral fica

    $\mathsf{\displaystyle=\int \frac{1}{10\,sec^2\,\theta}\cdot \sqrt{10}\,sec^2\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle=\frac{\sqrt{10}}{10}\int 1\,d\theta}\\\\\\ \mathsf{=\dfrac{\sqrt{10}}{10}\,\theta+C}\\\\\\ \mathsf{=\dfrac{\sqrt{10}}{10}\,arctg\!\left(\dfrac{2x}{\sqrt{10}}\right)+C\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)