Question Number 73202 by MJS last updated on 08/Nov/19
$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}=? \\ $$$$\int\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }}=? \\ $$
Commented by mathmax by abdo last updated on 09/Nov/19
$${let}\:{A}\:=\int\:\:\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{changement}\:{x}={cos}\mathrm{2}{t}\:{give} \\ $$$${A}\:=\int\:\:\frac{−\mathrm{2}{sin}\left(\mathrm{2}{t}\right)\:{dt}}{{cos}\left(\mathrm{2}{t}\right)\sqrt{\mathrm{2}{cos}^{\mathrm{2}} {t}}\left(\mathrm{2}{sin}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:=−\mathrm{2}\:\int\:\:\:\frac{{sin}\left(\mathrm{2}{t}\right){dt}}{{cos}\left(\mathrm{2}{t}\right)\sqrt{\mathrm{2}}{cost}\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{sin}^{\mathrm{3}} {t}} \\ $$$$=−\mathrm{2}\:\int\:\:\frac{\mathrm{2}{sint}\:{cost}}{\mathrm{4}\:{cos}\left(\mathrm{2}{t}\right){cost}\:{sint}\:{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$=−\int\:\:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)\left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)}\:=−\mathrm{2}\:\int\:\:\:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)−{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)} \\ $$$$=−\mathrm{2}\:\int\:\:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)−\frac{\mathrm{1}+{cos}\left(\mathrm{4}{t}\right)}{\mathrm{2}}}\:=−\mathrm{4}\:\int\:\:\:\frac{{dt}}{\mathrm{2}{cos}\left(\mathrm{2}{t}\right)−{cos}\left(\mathrm{4}{t}\right)−\mathrm{1}} \\ $$$$=_{\mathrm{2}{t}={z}} \:\:\:\:−\mathrm{4}\:\int\:\:\:\:\frac{{dz}}{\mathrm{2}\left(\mathrm{2}{cos}\left({z}\right)−{cos}\left(\mathrm{2}{z}\right)−\mathrm{1}\right)} \\ $$$$=−\mathrm{2}\:\int\:\:\:\:\frac{{dz}}{\mathrm{2}{cosz}\:−{cos}\left(\mathrm{2}{z}\right)−\mathrm{1}} \\ $$$$=−\mathrm{2}\:\int\:\:\frac{{dz}}{\mathrm{2}{cos}\left({z}\right)−\left(\mathrm{2}{cos}^{\mathrm{2}} {z}−\mathrm{1}\right)−\mathrm{1}} \\ $$$$=−\mathrm{2}\:\int\:\:\frac{{dz}}{−\mathrm{2}{cos}^{\mathrm{2}} {z}+\mathrm{2}{cosz}}\:=\int\:\:\frac{{dz}}{{cos}^{\mathrm{2}} {z}\:−{cosz}} \\ $$$$=\int\:\:\frac{{dz}}{{cosz}\left({cosz}−\mathrm{1}\right)}\:=\int\:\left(\frac{\mathrm{1}}{{cosz}−\mathrm{1}}−\frac{\mathrm{1}}{{cosz}}\right){dz} \\ $$$$=\int\:\:\frac{{dz}}{{cosz}−\mathrm{1}}−\int\:\:\frac{{dz}}{{cosz}}\:{we}\:{have} \\ $$$$\int\:\:\frac{{dz}}{{cosz}−\mathrm{1}}\:=_{{tan}\left(\frac{{z}}{\mathrm{2}}\right)={u}} \:\:\:\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }−\mathrm{1}}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} −\mathrm{1}−{u}^{\mathrm{2}} }\:=\int−\:\frac{{du}}{{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{u}}\:+{C}\:=\frac{\mathrm{1}}{{tan}\left(\frac{{z}}{\mathrm{2}}\right)}\:+{c} \\ $$$$=\frac{\mathrm{1}}{{tan}\left({t}\right)}\:+{C}\:=\frac{\mathrm{1}}{{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}\:+{C}\:\:{also} \\ $$$$\int\:\:\frac{{dz}}{{cosz}}\:=_{{tan}\left(\frac{{z}}{\mathrm{2}}\right)={u}} \:\:\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} }\:=\int\left(\frac{\mathrm{1}}{\mathrm{1}+{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du} \\ $$$$={ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\:+{c}\:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{z}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{z}}{\mathrm{2}}\right)}\mid\:+{c}\:={ln}\mid\frac{\mathrm{1}+{tan}\left({t}\right)}{\mathrm{1}−{tan}\left({t}\right)}\mid\:+{c} \\ $$$$={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}\mid\:+{c}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}\:−{ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{arcosx}}{\mathrm{2}}\right)}\mid\:+{C}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MJS last updated on 10/Nov/19
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
$${you}\:{are}\:{welcome}. \\ $$
Answered by MJS last updated on 10/Nov/19
$$\int\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:\rightarrow\:{dx}=\sqrt{{x}+\mathrm{1}}\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}= \\ $$$$=\int{dt}+\int\frac{{dt}}{{t}−\mathrm{1}}−\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$={t}+\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:=\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{\mathrm{1}−{x}}}+\mathrm{ln}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+{C} \\ $$
Answered by MJS last updated on 10/Nov/19
$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dt}\right] \\ $$$$=\int\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}{dt}={t}+\mathrm{2ln}\:\left({t}−\mathrm{1}\right)\:= \\ $$$$={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{2ln}\:\mid{x}−\mathrm{1}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid\:+{C} \\ $$