Question Number 176180 by gloriousman last updated on 14/Sep/22
$$\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}} \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 14/Sep/22
$${I}=\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}=\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\:\:=\mathrm{sgn}\left({x}+\mathrm{1}\right)\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}}},\:{t}=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow{dt}=−\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=−\mathrm{sgn}\left({t}\right)\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}}}=−\mathrm{sgn}\left({t}\right)\left(\mathrm{2}\sqrt{\mathrm{1}−{t}}\right)+{C} \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{sgn}\left({x}+\mathrm{1}\right)\left(\mathrm{2}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}}\right)+{C}=−\mathrm{sgn}\left({x}+\mathrm{1}\right)\left(\frac{\mathrm{2}\sqrt{{x}}}{\:\sqrt{{x}+\mathrm{1}}}\right)+{C} \\ $$
Answered by qaz last updated on 14/Sep/22
$${y}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\:\:\:{z}={x}+\mathrm{1}\:\:\:\Rightarrow{y}^{\mathrm{2}} ={z}^{\mathrm{2}} −\mathrm{1}\:\:\:{ydy}={zdz} \\ $$$$\Rightarrow{I}=\int\frac{{dz}}{{zy}}=\int\frac{{dy}}{{z}^{\mathrm{2}} }=\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{y}+{c} \\ $$$$=\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}+{c} \\ $$
Answered by BaliramKumar last updated on 15/Sep/22
$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}\:=\:\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}}\: \\ $$$${let}\:\:\:\left({x}+\mathrm{1}\right)\:=\:{sec}\theta,\:\:\:\:\:\theta\:=\:{sec}^{−\mathrm{1}} \left({x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:{dx}\:=\:{sec}\theta\centerdot{tan}\theta\centerdot{d}\theta \\ $$$$\int\frac{{sec}\theta\centerdot{tan}\theta\centerdot{d}\theta}{{sec}\theta\sqrt{{sec}^{\mathrm{2}} \theta\:−\:\mathrm{1}}}\:=\:\int\frac{{sec}\theta\centerdot{tan}\theta\centerdot{d}\theta}{{sec}\theta\centerdot{tan}\theta}\:=\: \\ $$$$\int\mathrm{1}{d}\theta\:=\:\theta\:+\:{C} \\ $$$$\:\:\:=\:\:{sec}^{−\mathrm{1}} \left({x}\:+\:\mathrm{1}\right)\:+\:\mathrm{C}\: \\ $$$$ \\ $$$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}\:=\:{tan}^{−\mathrm{1}} \left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\right)\:+\:{C} \\ $$$${both}\:{are}\:{right} \\ $$
Answered by BaliramKumar last updated on 14/Sep/22
$$ \\ $$$$\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}{dx} \\ $$$${let}\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\:=\:{y},\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{x}\:=\:{y}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}\left(\mathrm{2}{x}+\mathrm{2}\right){dx}\:=\:{dy} \\ $$$$\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}\:=\:\frac{{dy}}{\left({x}+\mathrm{1}\right)} \\ $$$$\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}{dx}\:=\:\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}\centerdot\frac{{dy}}{\left({x}+\mathrm{1}\right)}\: \\ $$$$\int\frac{{dy}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\int\frac{{dy}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)}\: \\ $$$$=\:\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{tan}^{−\mathrm{1}} \left({y}\right)\:+\:\mathrm{C} \\ $$$$=\:{tan}^{−\mathrm{1}} \left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\right)\:+\:{C} \\ $$
Commented by gloriousman last updated on 15/Sep/22
$$ \\ $$$$\mathrm{Many}\:\mathrm{Thanks}..\mathrm{Sir} \\ $$