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Question Number 85676 by john santu last updated on 24/Mar/20

$$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\$$ $${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\$$ $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\$$ $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{sin}\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{t}\right)^{\mathrm{4}} } \\$$ $$=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\mathrm{4cos}^{\mathrm{4}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)} \\$$ $$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)\:{dt} \\$$ $$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}=\:{u}\right] \\$$ $$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} {u}\:×\mathrm{2}{du} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\:}}\left(\mathrm{tan}\:^{\mathrm{2}} {u}+\mathrm{1}\right)\:{d}\left(\mathrm{tan}\:{u}\right) \\$$ $$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} {u}\:+\:\mathrm{tan}\:{u}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \\$$ $$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{0}−\left(−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\right]=\:\frac{\mathrm{2}}{\mathrm{3}} \\$$ $$\\$$

Commented byjagoll last updated on 24/Mar/20

$$\mathrm{i}\:\mathrm{try}\:\mathrm{by}\:\:\mathrm{Euler}\:\mathrm{substitution} \\$$ $$\mathrm{let}\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{x}+\mathrm{t}\: \\$$ $$\mathrm{1}+\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2xt}+\mathrm{t}^{\mathrm{2}} \\$$ $$\mathrm{2xt}+\mathrm{t}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:\mathrm{x}\:=\:\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}} \\$$ $$\mathrm{dx}\:=\:\frac{−\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2t}^{\mathrm{2}} }\:\mathrm{dt} \\$$ $$\int\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}}+\frac{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}}\right)^{\mathrm{2}} }\:×\:\left(\frac{−\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2t}^{\mathrm{2}} }\right)\:\mathrm{dt} \\$$ $$\int\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}^{\mathrm{2}} }\:×\left(−\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\:\mathrm{dt}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{dt} \\$$ $$−\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{3}} +\mathrm{t}\:\right]\:=−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{t}\:\left[\mathrm{t}^{\mathrm{2}} +\mathrm{3}\:\right] \\$$ $$−\frac{\mathrm{1}}{\mathrm{6}}\:\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:−\mathrm{x}\:\right)\:\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{4}−\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right) \\$$ $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:−\frac{\mathrm{1}}{\mathrm{6}}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{4}−\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\:+\frac{\mathrm{4}}{\mathrm{6}} \\$$ $$=\:\frac{\mathrm{2}}{\mathrm{3}} \\$$

Commented byjohn santu last updated on 24/Mar/20

$${good} \\$$

Commented bysakeefhasan05@gmail.com last updated on 24/Mar/20

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{n}} }\mathrm{dx}=\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} −\mathrm{1}} \\$$ $$\mathrm{common}\:\mathrm{solution} \\$$

Commented bysakeefhasan05@gmail.com last updated on 24/Mar/20

$$\mathrm{comment}\:\mathrm{pls} \\$$

Commented byjohn santu last updated on 24/Mar/20

$${waw}...\:{it}\:{generally}\:{solution}\:{sir}? \\$$

Commented bysakeefhasan05@gmail.com last updated on 24/Mar/20

$$\mathrm{yeah}\:.\mathrm{n}=\mathrm{3}\:\mathrm{check}\:\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\:\mathrm{pls}\:\mathrm{try} \\$$

Commented byjagoll last updated on 24/Mar/20

$$\mathrm{but}\:\mathrm{n}\:\neq\:\mathrm{1}? \\$$

Commented bysakeefhasan05@gmail.com last updated on 24/Mar/20

$$\mathrm{sry}\:\frac{\mathrm{n}}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{1}\right)}\:\:,\left[\mathrm{n}^{\mathrm{2}} −\mathrm{1}\neq\mathrm{0}\right]\:\mathrm{so}\:\:\left(\mathrm{n}\neq\mathrm{1}\right)\:\&\:\left(\mathrm{n}\neq−\mathrm{1}\right) \\$$

Commented bysakeefhasan05@gmail.com last updated on 24/Mar/20

$$\mathrm{thank}\:\mathrm{you}\:\:\mathrm{for}\:\mathrm{direct}\:\mathrm{me} \\$$

Commented bymathmax by abdo last updated on 24/Mar/20

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}} }\:\:{we}\:{do}\:{the}\:{changement}\:{x}={sh}\left({t}\right)\:\Rightarrow \\$$ $${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left({t}\right)}{\left({sh}\left({t}\right)+{ch}\left({t}\right)\right)^{{n}} }{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left({t}\right)}{\left(\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}+\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right)^{{n}} }{dt}\:\: \\$$ $$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{nt}} \left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\left(\:{e}^{\left(−{n}+\mathrm{1}\right){t}} \:+{e}^{−\left({n}+\mathrm{1}\right){t}} \right){dt} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}−{n}}{e}^{\left(\mathrm{1}−{n}\right){t}} \:−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\left({n}+\mathrm{1}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\mathrm{1}−{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{n}}{{n}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\$$ $${A}_{{n}} =\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{1}}\:\:\:\left({n}>\mathrm{1}\right)\:\:{so} \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\$$

Commented byjohn santu last updated on 24/Mar/20

$${thank}\:{you}\:{sir}\:{for}\:{your}\:{short}\:{cut} \\$$

Commented byjohn santu last updated on 24/Mar/20

$${indonesian}\:{sir} \\$$

Commented bymathmax by abdo last updated on 24/Mar/20

$${where}\:{are}\:{you}\:{from}\:{sir}\:{john}... \\$$

Commented bymathmax by abdo last updated on 24/Mar/20

$${aah}\:{good}\:\:{sir}\:... \\$$

Commented byjohn santu last updated on 25/Mar/20

$${why}\:{sir}?\:{if}\:{you}\:{where}\:{you}\: \\$$ $${come}\:{sir}? \\$$

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