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Question Number 38123 by maxmathsup by imad last updated on 22/Jun/18
f is a function positive and C^1 1) find ∫ (f^′ /(2(√f)(√(1+f))))dx 2)let A_n = ∫_0 ^1 (x^(n/2) /(x(√(1+x^n )))) calculate A_n and lim_(n→+∞) A_n
$${f}\:{is}\:{a}\:{function}\:{positive}\:\:{and}\:{C}^{\mathrm{1}} \:\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\:\:\frac{{f}^{'} }{\mathrm{2}\sqrt{{f}}\sqrt{\mathrm{1}+{f}}}{dx} \\ $$$$\left.\mathrm{2}\right){let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}^{\frac{{n}}{\mathrm{2}}} }{{x}\sqrt{\mathrm{1}+{x}^{{n}} }} \\ $$$${calculate}\:{A}_{{n}} \:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 22/Jun/18
changement (√f)=u (u function!) give ∫ (f^′ /(2(√f)(√(1+f))))dx=∫ ((2u^′ u)/(2u(√(1+u^2 ))))dx =∫ (u^′ /( (√(1+u^2 ))))dx=ln(u +(√(1+u^2 ))) +c =ln((√f) +(√(1+f))) +c 2) we have A_n = ∫_0 ^1 (x^(n/2) /(x(√(1+x^n ))))dx changement x^n =t give x=t^(1/n) and A_n = ∫_0 ^1 (t^(1/2) /(t^((1/n) ) (√(1+t)))) (1/n)t^((1/n) −1) dt =(1/n) ∫_0 ^1 (dt/( (√t)(√(1+t)))) =_((√t)=u) (1/n)∫_0 ^1 ((2udu)/(u(√(1+u^2 )))) = (2/n)[ln(u +(√(1+u^2 )))]_0 ^1 =(2/n){ ln(1+(√2))}
$${changement}\:\sqrt{{f}}={u}\:\left({u}\:{function}!\right)\:{give} \\ $$$$\int\:\:\frac{{f}^{'} }{\mathrm{2}\sqrt{{f}}\sqrt{\mathrm{1}+{f}}}{dx}=\int\:\frac{\mathrm{2}{u}^{'} {u}}{\mathrm{2}{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{dx} \\ $$$$=\int\:\:\:\:\frac{{u}^{'} }{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{dx}={ln}\left({u}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\:+{c} \\ $$$$={ln}\left(\sqrt{{f}}\:+\sqrt{\mathrm{1}+{f}}\right)\:+{c}\:\: \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}^{\frac{{n}}{\mathrm{2}}} }{{x}\sqrt{\mathrm{1}+{x}^{{n}} }}{dx}\: \\ $$$${changement}\:{x}^{{n}} ={t}\:{give}\:{x}={t}^{\frac{\mathrm{1}}{{n}}} \:\:{and} \\ $$$${A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{t}^{\frac{\mathrm{1}}{{n}}\:} \sqrt{\mathrm{1}+{t}}}\:\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+{t}}}\:=_{\sqrt{{t}}={u}} \:\:\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{udu}}{{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{2}}{{n}}\left[{ln}\left({u}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{{n}}\left\{\:{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
1)∫((df/dx)/(2(√f) (√(1+f))))dx ∫(df/(2(√f) (√(1+f)))) t^2 =1+f ∫((2tdt)/(2(√(t^2 −1)) ×t)) ∫(dt/( (√(t^2 −1)))) use formula ∫(dx/( (√(x^2 −a^2 )))) ln(t+(√(t^2 −1)) )+c ln((√(1+f)) +(√f) )+c
$$\left.\mathrm{1}\right)\int\frac{\frac{{df}}{{dx}}}{\mathrm{2}\sqrt{{f}}\:\sqrt{\mathrm{1}+{f}}}{dx} \\ $$$$\int\frac{{df}}{\mathrm{2}\sqrt{{f}}\:\sqrt{\mathrm{1}+{f}}} \\ $$$${t}^{\mathrm{2}} =\mathrm{1}+{f} \\ $$$$\int\frac{\mathrm{2}{tdt}}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:×{t}} \\ $$$$\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:\:{use}\:{formula}\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c} \\ $$$${ln}\left(\sqrt{\mathrm{1}+{f}}\:+\sqrt{{f}}\:\right)+{c} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
x^n +1=t^2 2tdt =nx^(n−1) dx 2tdt=nx^n (dx/x) ((2tdt)/(n(t^2 −1)))=(dx/x) ∫_1 ^((√2) ) ((2tdt)/(n(t^2 −1)))×(((√(t^2 −1)) )/t) =(2/n)∫_1 ^(√2) (√(t^2 −1)) dt (1/n)∣(t/2)(√(t^2 −1)) −(1/2)ln(t+(√(t^2 −1))) ∣_1 ^((√2) ) (1/n){(((√2) )/2)−(1/2)ln((√2) +1)} =(1/n){(((√2) )/2)−(1/2)ln((√2) +1)} when n→∞ value of intregation is 0 use formula∫(√(x^2 −a^2 )) dx
$${x}^{{n}} +\mathrm{1}={t}^{\mathrm{2}} \:\:\:\:\mathrm{2}{tdt}\:={nx}^{{n}−\mathrm{1}} {dx} \\ $$$$ \\ $$$$\mathrm{2}{tdt}={nx}^{{n}} \frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{2}{tdt}}{{n}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{dx}}{{x}} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}\:} \frac{\mathrm{2}{tdt}}{{n}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}×\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:}{{t}} \\ $$$$=\frac{\mathrm{2}}{{n}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:{dt} \\ $$$$\frac{\mathrm{1}}{{n}}\mid\frac{{t}}{\mathrm{2}}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}+\sqrt{\left.{t}^{\mathrm{2}} −\mathrm{1}\right)}\:\mid_{\mathrm{1}} ^{\sqrt{\mathrm{2}}\:} \right. \\ $$$$\frac{\mathrm{1}}{{n}}\left\{\frac{\sqrt{\mathrm{2}}\:}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{n}}\left\{\frac{\sqrt{\mathrm{2}}\:}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\right\} \\ $$$${when}\:{n}\rightarrow\infty\:\:{value}\:{of}\:{intregation}\:{is}\:\mathrm{0} \\ $$$${use}\:{formula}\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\:{dx} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

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