Question Number 162055 by Astro last updated on 25/Dec/21
$$\int{e}^{\mathrm{2x}} \sqrt{\left(\mathrm{1}\:−{e}^{\mathrm{2}{x}} \right)}{dx} \\ $$
Answered by aleks041103 last updated on 25/Dec/21
$${u}=\mathrm{1}−{e}^{\mathrm{2}{x}} \\ $$$${du}=−\mathrm{2}{e}^{\mathrm{2}{x}} {dx}\Rightarrow{e}^{\mathrm{2}{x}} {dx}=−\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\Rightarrow\int{e}^{\mathrm{2}{x}} \sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{u}}{du}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}{u}\sqrt{{u}} \\ $$$$\Rightarrow\int{e}^{\mathrm{2}{x}} \sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }{dx}=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\mathrm{3}/\mathrm{2}} +{C} \\ $$