Question Number 153563 by ZiYangLee last updated on 08/Sep/21
$$\int\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}\:}\:{dx}\:=? \\ $$
Answered by puissant last updated on 08/Sep/21
$${u}=\sqrt{\mathrm{2}{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{2}{x}\:\rightarrow\:{x}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:{dx}={udu} \\ $$$${K}=\int\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}}{dx}=\int\frac{{u}}{\mathrm{1}+{u}}{du} \\ $$$$=\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$$={u}−{ln}\mid\mathrm{1}+{u}\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:\:{K}=\sqrt{\mathrm{2}{x}}−{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}{x}}\mid+{C}.. \\ $$