Question Number 215134 by RoseAli last updated on 29/Dec/24
Answered by Ghisom last updated on 29/Dec/24
![∫(√(x−(√(x^2 −4))))dx= [t=(√(x−(√(x^2 −4)))) → dx=−((2(√(x^2 −4)))/t)dt] =∫(t^2 −(4/t^2 ))dt=(t^3 /3)+(4/t)= =((2(2x+(√(x^2 −4))))/3)(√(x−(√(x^2 −4))))+C](https://www.tinkutara.com/question/Q215135.png)
$$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{{t}}{dt}\right] \\ $$$$=\int\left({t}^{\mathrm{2}} −\frac{\mathrm{4}}{{t}^{\mathrm{2}} }\right){dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4}}{{t}}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)}{\mathrm{3}}\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}+{C} \\ $$
Answered by MathematicalUser2357 last updated on 30/Dec/24
Commented by Frix last updated on 31/Dec/24
$$\mathrm{Wrong}. \\ $$
Answered by MrGaster last updated on 01/Jan/25
$$−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}−\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\mid\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}+\sqrt{\left(\mathrm{3}−\sqrt{\left.{x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }+\mathrm{4}\right.}\mid+{C}\right) \\ $$
Commented by Frix last updated on 01/Jan/25
$$\mathrm{WTF}? \\ $$