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Question Number 85162 by mathmax by abdo last updated on 19/Mar/20

1)find ∫ ln((√x)+(√(x+1)))dx 2) calculate ∫_0 ^1 ln((√x)+(√(x+1)))dx

$$\left.\mathrm{1}\right){find}\:\int\:{ln}\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$

Commented by mathmax by abdo last updated on 01/Apr/20

1) let f(t) =∫ ln(t+(√x) +(√(x+1)))dx we have f^′ (t) ∫ (1/(t+(√x)+(√(x+1))))dx =_((√x)=u) ∫ ((2udu)/(t+u+(√(u^2 +1)))) =_(u =sh(α)) 2 ∫ ((sh(α)ch(α)dα)/(t+sh(α)+ch(α))) = ∫ ((sh(2α))/(t+shα +chα))dα =∫ (((e^(2α) −e^(−α) )/2)/(t+ e^α ))dα =_(e^α =z) (1/2)∫ ((z^2 −z^(−2) )/(t+z)) (dz/z) =(1/2)∫ ((z^2 −z^(−2) )/(z(t+z)))dz =(1/2)∫ ((z^4 −1)/(z^3 (t+z)))dz let decompose F(z) =((z^4 −1)/(z^3 (z+t)))⇒F(z) =((z^4 −1)/(z^4 +tz^3 )) =((z^4 +tz^3 −tz^3 −1)/(z^4 +tz^3 )) =1−((tz^3 +1)/(z^4 +tz^3 )) w(z)=((tz^(3 ) +1)/(z^3 (z+t))) =(a/z)+(b/z^2 )+(c/z^3 ) +(d/(z+t)) c =(1/t) , d =((1−t^4 )/(−t^3 )) =((t^4 −1)/t^3 ) ⇒w(z) =(a/z)+(b/z^2 ) +(1/(tz^3 )) +((t^4 −1)/(t^3 (z+t))) lim_(z→+∞) zw(z) =t =a +d ⇒a =t−d =t−((t^4 −1)/t^3 ) =(1/t^3 ) ⇒ w(z) =(1/(t^3 z)) +(b/z^2 ) +(1/(tz^3 )) +((t^4 −1)/(t^3 (z+t))) w(1) =1 =(1/t^3 ) +b +(1/t) +((t^4 −1)/(t+1)) =(1/t^3 )+(1/t) +((t^4 −1)/(t+1)) =((t+t^3 )/t^4 ) +((t^4 −1)/(t+1)) =(((t+t^3 )(t+1)+t^8 −t^4 )/(t^4 (t+1))) =((t^2 +t+t^4 +t^3 +t^8 −t^4 )/(t^4 (t+1))) =((t^8 +t^3 +t^2 +t)/(t^4 (t+1))) =((t^7 +t^2 +t+1)/(t^3 (t+1))) ⇒ b=1−((t^7 +t^2 +t+1)/(t^3 (t+1))) we have F(z)=1−(a/z)−(b/z^2 )−(c/z^3 )−(d/(z+t)) ⇒ ∫ F(z)dz =z−aln∣z∣+(b/z) +(c/(2z^2 )) −d ln∣z+t∣ +C...be continued...

$$\left.\mathrm{1}\right)\:{let}\:{f}\left({t}\right)\:=\int\:{ln}\left({t}+\sqrt{{x}}\:+\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$$${we}\:{have}\:{f}^{'} \left({t}\right)\:\int\:\:\:\frac{\mathrm{1}}{{t}+\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}{dx}\:=_{\sqrt{{x}}={u}} \:\:\int\:\:\:\frac{\mathrm{2}{udu}}{{t}+{u}+\sqrt{{u}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$$=_{{u}\:={sh}\left(\alpha\right)} \:\:\mathrm{2}\:\int\:\:\frac{{sh}\left(\alpha\right){ch}\left(\alpha\right){d}\alpha}{{t}+{sh}\left(\alpha\right)+{ch}\left(\alpha\right)}\:=\:\int\:\:\:\frac{{sh}\left(\mathrm{2}\alpha\right)}{{t}+{sh}\alpha\:+{ch}\alpha}{d}\alpha \\ $$$$=\int\:\:\frac{\frac{{e}^{\mathrm{2}\alpha} −{e}^{−\alpha} }{\mathrm{2}}}{{t}+\:{e}^{\alpha} }{d}\alpha\:\:=_{{e}^{\alpha} ={z}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{z}^{\mathrm{2}} −{z}^{−\mathrm{2}} }{{t}+{z}}\:\frac{{dz}}{{z}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{z}^{\mathrm{2}} −{z}^{−\mathrm{2}} }{{z}\left({t}+{z}\right)}{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{z}^{\mathrm{4}} −\mathrm{1}}{{z}^{\mathrm{3}} \left({t}+{z}\right)}{dz}\:\:{let}\:{decompose} \\ $$$${F}\left({z}\right)\:=\frac{{z}^{\mathrm{4}} −\mathrm{1}}{{z}^{\mathrm{3}} \left({z}+{t}\right)}\Rightarrow{F}\left({z}\right)\:=\frac{{z}^{\mathrm{4}} −\mathrm{1}}{{z}^{\mathrm{4}} +{tz}^{\mathrm{3}} }\:=\frac{{z}^{\mathrm{4}} +{tz}^{\mathrm{3}} −{tz}^{\mathrm{3}} −\mathrm{1}}{{z}^{\mathrm{4}} \:+{tz}^{\mathrm{3}} }\:=\mathrm{1}−\frac{{tz}^{\mathrm{3}} \:+\mathrm{1}}{{z}^{\mathrm{4}} +{tz}^{\mathrm{3}} } \\ $$$${w}\left({z}\right)=\frac{{tz}^{\mathrm{3}\:} +\mathrm{1}}{{z}^{\mathrm{3}} \left({z}+{t}\right)}\:=\frac{{a}}{{z}}+\frac{{b}}{{z}^{\mathrm{2}} }+\frac{{c}}{{z}^{\mathrm{3}} }\:+\frac{{d}}{{z}+{t}} \\ $$$${c}\:=\frac{\mathrm{1}}{{t}}\:\:,\:\:\:\:{d}\:=\frac{\mathrm{1}−{t}^{\mathrm{4}} }{−{t}^{\mathrm{3}} }\:=\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}^{\mathrm{3}} }\:\Rightarrow{w}\left({z}\right)\:=\frac{{a}}{{z}}+\frac{{b}}{{z}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{tz}^{\mathrm{3}} }\:+\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}^{\mathrm{3}} \left({z}+{t}\right)} \\ $$$${lim}_{{z}\rightarrow+\infty} \:{zw}\left({z}\right)\:={t}\:={a}\:+{d}\:\Rightarrow{a}\:={t}−{d}\:={t}−\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:\Rightarrow \\ $$$${w}\left({z}\right)\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} {z}}\:+\frac{{b}}{{z}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{tz}^{\mathrm{3}} }\:+\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}^{\mathrm{3}} \left({z}+{t}\right)} \\ $$$${w}\left(\mathrm{1}\right)\:=\mathrm{1}\:\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:+{b}\:+\frac{\mathrm{1}}{{t}}\:+\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}+\mathrm{1}}\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{{t}}\:+\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}+\mathrm{1}} \\ $$$$=\frac{{t}+{t}^{\mathrm{3}} }{{t}^{\mathrm{4}} }\:+\frac{{t}^{\mathrm{4}} −\mathrm{1}}{{t}+\mathrm{1}}\:=\frac{\left({t}+{t}^{\mathrm{3}} \right)\left({t}+\mathrm{1}\right)+{t}^{\mathrm{8}} −{t}^{\mathrm{4}} }{{t}^{\mathrm{4}} \left({t}+\mathrm{1}\right)} \\ $$$$=\frac{{t}^{\mathrm{2}} \:+{t}+{t}^{\mathrm{4}} +{t}^{\mathrm{3}} +{t}^{\mathrm{8}} −{t}^{\mathrm{4}} }{{t}^{\mathrm{4}} \left({t}+\mathrm{1}\right)}\:=\frac{{t}^{\mathrm{8}} +{t}^{\mathrm{3}} +{t}^{\mathrm{2}} \:+{t}}{{t}^{\mathrm{4}} \left({t}+\mathrm{1}\right)}\:=\frac{{t}^{\mathrm{7}} \:+{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}{{t}^{\mathrm{3}} \left({t}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${b}=\mathrm{1}−\frac{{t}^{\mathrm{7}} \:+{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}{{t}^{\mathrm{3}} \left({t}+\mathrm{1}\right)}\:{we}\:{have}\:{F}\left({z}\right)=\mathrm{1}−\frac{{a}}{{z}}−\frac{{b}}{{z}^{\mathrm{2}} }−\frac{{c}}{{z}^{\mathrm{3}} }−\frac{{d}}{{z}+{t}}\:\Rightarrow \\ $$$$\int\:{F}\left({z}\right){dz}\:={z}−{aln}\mid{z}\mid+\frac{{b}}{{z}}\:+\frac{{c}}{\mathrm{2}{z}^{\mathrm{2}} }\:−{d}\:{ln}\mid{z}+{t}\mid\:+{C}...{be}\:{continued}... \\ $$

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