Question Number 54896 by yusufbode1996 last updated on 14/Feb/19
$$\int\sqrt{}\left({x}\mathrm{2}+\mathrm{2}{x}+\mathrm{2}\right) \\ $$
Commented by maxmathsup by imad last updated on 14/Feb/19
$${let}\:{I}\:=\int\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx}\:\Rightarrow{I}\:=\int\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:{changement}\:{x}+\mathrm{1}\:={sh}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:{ch}\left({t}\right){ch}\left({t}\right)\:{dt}\:=\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{t}}{\mathrm{2}}\:\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{argsh}\left({x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\right)\:+\frac{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}}{\mathrm{2}}\:+{c}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19
![∫(√(x^2 +2x+1+1)) dx ∫(√((x+1)^2 +1))dx now x+1=y 1=a^2 ∫(√(y^2 +a^2 )) dy (√(y^2 +a^2 )) ∫dy−∫[(d/dy)(√(y^2 +a^2 )) ∫dy]dy y(√(y^2 +a^2 )) −∫((2y×y)/(2(√(y^2 +a^2 ))))dy y(√(y^2 +a^2 )) −∫((y^2 +a^2 −a^2 )/(√(y^2 +a^2 )))dy I=y(√(y^2 +a^2 )) −∫(√(y^2 +a^2 )) dy+a^2 ∫(dy/(√(y^2 +a^2 ))) 2I=y(√(y^2 +a^2 )) +a^2 ln(y+(√(y^2 +a^2 )) ) I=(y/2)(√(y^2 +a^2 )) +(a^2 /2)ln(y+(√(y^2 +a^2 )) )+c so required answer is =(((x+1))/2)(√((x+1)^2 +1)) +(1/2)ln{(x+1)+(√((x+1)^2 +1)) }+c](Q54898.png)
$$\int\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+\mathrm{1}}\:{dx} \\ $$$$\int\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$ \\ $$$${now}\:{x}+\mathrm{1}={y}\:\:\:\:\mathrm{1}={a}^{\mathrm{2}} \\ $$$$\int\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:{dy} \\ $$$$\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\int{dy}−\int\left[\frac{{d}}{{dy}}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\int{dy}\right]{dy} \\ $$$${y}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:−\int\frac{\mathrm{2}{y}×{y}}{\mathrm{2}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{dy} \\ $$$${y}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:−\int\frac{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{dy} \\ $$$${I}={y}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:−\int\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:{dy}+{a}^{\mathrm{2}} \int\frac{{dy}}{\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{I}={y}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+{a}^{\mathrm{2}} {ln}\left({y}+\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right) \\ $$$${I}=\frac{{y}}{\mathrm{2}}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({y}+\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right)+{c} \\ $$$${so}\:{required}\:{answer}\:{is} \\ $$$$=\frac{\left({x}+\mathrm{1}\right)}{\mathrm{2}}\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({x}+\mathrm{1}\right)+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:\right\}+{c} \\ $$$$ \\ $$