Question

calcule a integral abaixo:

Answer 1

$\displaystyle\int\frac{\sqrt{e^{-3x}-2}}{e^{3x}}\,dx\\\\\\ =\int e^{-3x}\sqrt{e^{-3x}-2}\,dx\\\\\\ =\int\frac{1}{(-3)}\cdot (-3)e^{-3x}\sqrt{e^{-3x}-2}\,dx\\\\\\ =-\frac{1}{3}\int\sqrt{e^{-3x}-2}\cdot (-3)e^{-3x}\,dx~~~~~~\mathbf{(i)}$


Fazendo a seguinte substituição:

$e^{-3x}-2=u~~\Rightarrow~~-3e^{-3x}\,dx=du$


Substituindo em $\mathbf{(i)},$ a integral fica

$=\displaystyle -\frac{1}{3}\int\sqrt{u}\,du\\\\\\ =-\frac{1}{3}\int u^{1/2}\,du\\\\\\ =-\frac{1}{3}\cdot \dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =-\frac{1}{3}\cdot \dfrac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =-\frac{1}{3}\cdot \dfrac{2}{3}\,u^{3/2}+C\\\\\\ =-\frac{2}{9}\,u^{3/2}+C\\\\\\ =-\frac{2}{9}\big(e^{-3x}-2\big)^{3/2}+C$

$\therefore~~\boxed{\begin{array}{c}\displaystyle\int\dfrac{\sqrt{e^{-3x}-2}}{e^{3x}}\,dx=-\frac{2}{9}\big(e^{-3x}-2\big)^{3/2}+C \end{array}}$


Bons estudos! :-)