Question Number 44092 by peter frank last updated on 21/Sep/18
Commented by maxmathsup by imad last updated on 21/Sep/18
![let I = ∫_0 ^∞ (dx/((x+(√(1+x^2 ))))) changement x =sh(t) give I = ∫_0 ^(+∞) ((ch(t)dt)/((sh(t) +ch(t))^2 )) =∫_0 ^∞ ((ch(t))/((((e^t −e^(−t) )/2)+((e^t +e^(−t) )/2))^2 ))dt = ∫_0 ^∞ ((ch(t))/e^(2t) ) dt =∫_0 ^∞ e^(−2t) ((e^t +e^(−t) )/2)dt =(1/2)∫_0 ^∞ (e^(−t) +e^(−3t) )dt =(1/2)[−e^(−t) −(1/3)e^(−3t) ]_0 ^(+∞) = (1/2){(4/3)} ⇒ I =(2/3) .](https://www.tinkutara.com/question/Q44127.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}\:\:{changement}\:{x}\:={sh}\left({t}\right)\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{ch}\left({t}\right){dt}}{\left({sh}\left({t}\right)\:+{ch}\left({t}\right)\right)^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left({t}\right)}{\left(\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}+\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left({t}\right)}{{e}^{\mathrm{2}{t}} }\:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\mathrm{2}{t}} \:\:\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{−{t}} \:+{e}^{−\mathrm{3}{t}} \right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[−{e}^{−{t}} \:−\frac{\mathrm{1}}{\mathrm{3}}{e}^{−\mathrm{3}{t}} \right]_{\mathrm{0}} ^{+\infty} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{4}}{\mathrm{3}}\right\}\:\Rightarrow\:{I}\:=\frac{\mathrm{2}}{\mathrm{3}}\:. \\ $$