Question-215134

Question Number 215134 by RoseAli last updated on 29/Dec/24

Answered by Ghisom last updated on 29/Dec/24

$$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{{t}}{dt}\right] \\ $$$$=\int\left({t}^{\mathrm{2}} −\frac{\mathrm{4}}{{t}^{\mathrm{2}} }\right){dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4}}{{t}}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)}{\mathrm{3}}\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}+{C} \\ $$

Answered by MathematicalUser2357 last updated on 30/Dec/24

Commented by Frix last updated on 31/Dec/24

$$\mathrm{Wrong}. \\ $$

Commented by MathematicalUser2357 last updated on 07/Feb/25

did you mean crong?����������

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

$$\mathrm{WTF}? \\ $$

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