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Question Number 153563 by ZiYangLee last updated on 08/Sep/21

∫ (1/(1+(√(2x)) )) dx =?

$$\int\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}\:}\:{dx}\:=? \\ $$

Answered by puissant last updated on 08/Sep/21

u=(√(2x)) → u^2 =2x → x=(u^2 /2) ⇒ dx=udu  K=∫(1/(1+(√(2x))))dx=∫(u/(1+u))du  =∫(1−(1/(1+u)))du  =u−ln∣1+u∣+C    ∴∵  K=(√(2x))−ln∣1+(√(2x))∣+C..

$${u}=\sqrt{\mathrm{2}{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{2}{x}\:\rightarrow\:{x}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:{dx}={udu} \\ $$$${K}=\int\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}}{dx}=\int\frac{{u}}{\mathrm{1}+{u}}{du} \\ $$$$=\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$$={u}−{ln}\mid\mathrm{1}+{u}\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:\:{K}=\sqrt{\mathrm{2}{x}}−{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}{x}}\mid+{C}.. \\ $$

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