Menu Close

find-dx-1-chx-2-2-calculate-0-1-dx-1-chx-2-




Question Number 34285 by math khazana by abdo last updated on 03/May/18
find  ∫      (dx/((1+chx)^2 ))  2) calculate  ∫_0 ^1      (dx/((1+chx)^2 ))
$${find}\:\:\int\:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{chx}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{chx}\right)^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 06/May/18
let put  I  = ∫   (dx/((1+chx)^2 ))  I = ∫       (dx/(( 1+((e^x  +e^(−x) )/2))^2 ))  = ∫     (4/((2 + e^x  +e^(−x) )^2 ))dx  =_(e^x  =t)       ∫        (4/((2  +t   +(1/t))^2 )) (dt/t)  = 4 ∫         (dt/(t( 2t +t^2  +1)^2 )) t^2  dt  =4 ∫         (t/((t^2  +2t +1)^2 ))dt  =4 ∫       (t/((t+1)^4 )) dt  =4 ∫ ((t+1−1)/((t+1)^4 )) dt  = 4 ∫    (dt/((t+1)^3 )) −4 ∫    (dt/((t+1)^4 ))  =4 (1/(−3+1))(t+1)^(−3+1)    −4  (1/(−4+1))(t+1)^(−4+1)   =−2 (t+1)^(−2)    +(4/3) (t+1)^(−3)   = ((−2)/((t+1)^2 ))  +  (4/(3(t+1)^3 ))  I   =  ((−2)/((1 +e^x )^2 ))  +  (4/(3(1+e^x )^3 ))
$${let}\:{put}\:\:{I}\:\:=\:\int\:\:\:\frac{{dx}}{\left(\mathrm{1}+{chx}\right)^{\mathrm{2}} } \\ $$$${I}\:=\:\int\:\:\:\:\:\:\:\frac{{dx}}{\left(\:\mathrm{1}+\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}\right)^{\mathrm{2}} }\:\:=\:\int\:\:\:\:\:\frac{\mathrm{4}}{\left(\mathrm{2}\:+\:{e}^{{x}} \:+{e}^{−{x}} \right)^{\mathrm{2}} }{dx} \\ $$$$=_{{e}^{{x}} \:={t}} \:\:\:\:\:\:\int\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\left(\mathrm{2}\:\:+{t}\:\:\:+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} }\:\frac{{dt}}{{t}} \\ $$$$=\:\mathrm{4}\:\int\:\:\:\:\:\:\:\:\:\frac{{dt}}{{t}\left(\:\mathrm{2}{t}\:+{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{t}^{\mathrm{2}} \:{dt} \\ $$$$=\mathrm{4}\:\int\:\:\:\:\:\:\:\:\:\frac{{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{4}\:\int\:\:\:\:\:\:\:\frac{{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }\:{dt}\:\:=\mathrm{4}\:\int\:\frac{{t}+\mathrm{1}−\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }\:{dt} \\ $$$$=\:\mathrm{4}\:\int\:\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }\:−\mathrm{4}\:\int\:\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{4}\:\frac{\mathrm{1}}{−\mathrm{3}+\mathrm{1}}\left({t}+\mathrm{1}\right)^{−\mathrm{3}+\mathrm{1}} \:\:\:−\mathrm{4}\:\:\frac{\mathrm{1}}{−\mathrm{4}+\mathrm{1}}\left({t}+\mathrm{1}\right)^{−\mathrm{4}+\mathrm{1}} \\ $$$$=−\mathrm{2}\:\left({t}+\mathrm{1}\right)^{−\mathrm{2}} \:\:\:+\frac{\mathrm{4}}{\mathrm{3}}\:\left({t}+\mathrm{1}\right)^{−\mathrm{3}} \\ $$$$=\:\frac{−\mathrm{2}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{4}}{\mathrm{3}\left({t}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${I}\:\:\:=\:\:\frac{−\mathrm{2}}{\left(\mathrm{1}\:+{e}^{{x}} \right)^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} } \\ $$
Commented by prof Abdo imad last updated on 06/May/18
I = ((−2)/((1+e^x )^2 ))  + (4/(3( 1+e^x )^3 ))  + c  .
$${I}\:=\:\frac{−\mathrm{2}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }\:\:+\:\frac{\mathrm{4}}{\mathrm{3}\left(\:\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }\:\:+\:{c}\:\:. \\ $$
Commented by prof Abdo imad last updated on 06/May/18
∫_0 ^1       (dx/((1+chx)^2 ))  =[  (4/(3(1+e^x )^3 )) −(2/((1+e^x )^2 ))]_0 ^1   = (4/(3(1+e)^3 )) −(2/((1 +e)^2 )) −(4/(3.8)) +(2/2^2 )  = (4/(3( 1+e)^3 )) −(2/((1+e)^2 )) −(1/6) +(1/2)  = (4/(3(1+e)^3 )) −(2/((1+e)^2 ))  +(1/3) .
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{chx}\right)^{\mathrm{2}} }\:\:=\left[\:\:\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }\:−\frac{\mathrm{2}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{e}\right)^{\mathrm{3}} }\:−\frac{\mathrm{2}}{\left(\mathrm{1}\:+{e}\right)^{\mathrm{2}} }\:−\frac{\mathrm{4}}{\mathrm{3}.\mathrm{8}}\:+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}\left(\:\mathrm{1}+{e}\right)^{\mathrm{3}} }\:−\frac{\mathrm{2}}{\left(\mathrm{1}+{e}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{e}\right)^{\mathrm{3}} }\:−\frac{\mathrm{2}}{\left(\mathrm{1}+{e}\right)^{\mathrm{2}} }\:\:+\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *