Question Number 30769 by abdo imad last updated on 25/Feb/18
$${find}\:{the}\:{value}\:{of}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}}\:. \\ $$
Commented by abdo imad last updated on 27/Feb/18
$${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\:{the}\:{ch}={x}+\mathrm{1}={t}\:{give} \\ $$$${I}=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:{the}\:{we}\:{can}\:{use}\:{the}\:{ch}.{t}={tan}\theta\:\Rightarrow \\ $$$${I}=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{artan}\mathrm{2}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{{tan}^{\mathrm{2}} \theta\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}{d}\theta\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \:\:\:\frac{{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{2}} \theta}\:\frac{\mathrm{1}}{{cos}\theta}{d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \:\:\:\:\frac{{cos}\theta}{{sin}^{\mathrm{2}} \theta}\:{d}\theta\:\:=\left[−\frac{\mathrm{1}}{{sin}\theta}\right]_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}\:−\:\frac{\mathrm{1}}{{sin}\left({arctan}\mathrm{2}\right)}\:{but}\:{we}\:{have}\:{the}\:{formula} \\ $$$${sin}\left({arctanx}\right)=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\left(\:{for}\:{proof}\:{put}\:{x}={tan}\theta\right)\Rightarrow \\ $$$${sin}\left({arctan}\mathrm{2}\right)=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\Rightarrow\:{I}=\sqrt{\mathrm{2}}\:\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:. \\ $$