Integration Questions

Question Number 88007 by Mikael_786 last updated on 07/Apr/20

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}=? \\$$

Answered by MJS last updated on 07/Apr/20

$${t}=\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} \sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dt} \\$$ $$\Rightarrow \\$$ $$\int\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}=−\mathrm{16}\int\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dt}= \\$$ $$\:\:\:\:\:\left[\mathrm{Ostrogradski}\:\mathrm{again}\right] \\$$ $$=\frac{\mathrm{4}{t}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}}−\mathrm{4}\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}} \\$$ $$\\$$ $$\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{2}}{dt}+\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{2}}{dt}= \\$$ $$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{2}\right)\:+\frac{\sqrt{\mathrm{6}}}{\mathrm{24}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)}{\mathrm{3}}\:+ \\$$ $$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{2}\right)\:+\frac{\sqrt{\mathrm{6}}}{\mathrm{24}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{\mathrm{3}}\:= \\$$ $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{2}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{2}}\:+\frac{\sqrt{\mathrm{6}}}{\mathrm{24}}\left(\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{\mathrm{3}}\right) \\$$ $$\\$$ $$\Rightarrow \\$$ $$\int\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}= \\$$ $$=\frac{\mathrm{4}{t}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{2}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{2}}\:−\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\left(\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{\mathrm{3}}\right)\:+{C} \\$$ $$\mathrm{with}\:{t}=\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}} \\$$ $$\\$$ $$\Rightarrow \\$$ $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}=\frac{\pi}{\sqrt{\mathrm{6}}} \\$$

Commented byMikael_786 last updated on 08/Apr/20

$${thankfull}\:{Bro} \\$$