Question Number 36997 by rahul 19 last updated on 07/Jun/18 | ||
$$\int\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}\:}{dx}\:=\:? \\ $$ | ||
Commented by behi83417@gmail.com last updated on 07/Jun/18 | ||
$${x}={tg}^{\mathrm{2}} {t}\Rightarrow{dx}=\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt} \\ $$$${I}=\int\sqrt{\frac{\mathrm{1}+{tg}^{\mathrm{2}} {t}}{{tg}^{\mathrm{2}} {t}}.}\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}= \\ $$$$=\int\frac{\frac{\mathrm{1}}{{cost}}}{{tgt}}.\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}=\mathrm{2}\int{sec}^{\mathrm{3}} {tdt}= \\ $$$$={sect}.{tgt}+{ln}\left({sect}+{tgt}\right)+{C}= \\ $$$$=\sqrt{{x}\left(\mathrm{1}+{x}\right)}+{ln}\left(\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}\right)+{C}\:.\blacksquare \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18 | ||
$$\int\frac{\mathrm{1}+{x}}{\sqrt{{x}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{1}+\mathrm{2}{x}}{\sqrt{{x}+{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}.{x}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+{x}^{\mathrm{2}} \right)}{\sqrt{{x}+{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${use}\:{formula}\:\int\frac{{dx}}{\sqrt{{X}^{\mathrm{2}} −{A}^{\mathrm{2}} }} \\ $$ | ||
Answered by MJS last updated on 07/Jun/18 | ||
$$\int\sqrt{\frac{\mathrm{1}+{x}}{{x}}}{dx}=\int\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{{x}}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{arctan}\:{t}\:\rightarrow\:{dt}=\mathrm{sec}^{\mathrm{2}} \:{u}\:{du}\right] \\ $$$$=\mathrm{2}\int\mathrm{sec}^{\mathrm{2}} \:{u}\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{u}}\:{du}=\mathrm{2}\int\mathrm{sec}^{\mathrm{3}} \:{u}\:{du}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\int\mathrm{sec}^{{n}} \:{u}\:{du}=\frac{\mathrm{sec}^{{n}−\mathrm{2}} \:{u}\:\mathrm{tan}\:{u}}{{n}−\mathrm{1}}+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{sec}^{{n}−\mathrm{2}} \:{u}\:{du}\right] \\ $$$$=\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:+\int\mathrm{sec}\:{u}\:{du}= \\ $$$$=\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:+\mathrm{ln}\left(\mathrm{sec}\:{u}\:+\mathrm{tan}\:{u}\right)= \\ $$$$={t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{ln}\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+{t}\right)= \\ $$$$=\sqrt{{x}\left(\mathrm{1}+{x}\right)}+\mathrm{ln}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)+{C} \\ $$ | ||