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Question Number 40127 by maxmathsup by imad last updated on 16/Jul/18
find the value of ∫_0 ^1 ((e^x −1)/(e^x +1))dx
findthevalueof∫01ex−1ex+1dx
Commented by math khazana by abdo last updated on 18/Jul/18
let I = ∫_0 ^1 ((e^x −1)/(e^x +1))dx I = ∫_0 ^1 (e^x /(e^x +1))dx −∫_0 ^1 (dx/(e^x +1)) but ∫_0 ^1 (e^x /(e^x +1))dx =[ln(e^x +1)]_0 ^1 =ln(1+e)−ln(2) changement e^x =t give ∫_0 ^1 (dx/(e^x +1))dx = ∫_1 ^e (1/(t+1)) (dt/t) =∫_1 ^e ((1/t) −(1/(t+1)))dt =[ln∣ (t/(t+1))∣]_1 ^e = ln((e/(e+1))) +ln(2) ⇒ I =ln(1+e)−ln(2) −ln((e/(e+1)))−ln(2) =2ln(e+1) −2ln(2) −1 .
letI=∫01ex−1ex+1dxI=∫01exex+1dx−∫01dxex+1but∫01exex+1dx=[ln(ex+1)]01=ln(1+e)−ln(2)changementex=tgive∫01dxex+1dx=∫1e1t+1dtt=∫1e(1t−1t+1)dt=[ln∣tt+1∣]1e=ln(ee+1)+ln(2)⇒I=ln(1+e)−ln(2)−ln(ee+1)−ln(2)=2ln(e+1)−2ln(2)−1.
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18
∫_0 ^1 ((e^x −1)/(e^x +1))dx t=e^x +1 dt=e^x dx (dt/(t−1))=dx ∫_2 ^(e+1) ((t−2)/t).(dt/(t−1)) ∫_2 ^(e+1) ((2(t−1)−t)/(t(t−1)))dt ∫_2 ^(e+1) ((2dt)/t)−∫_2 ^(e+1) (dt/(t−1)) 2lnt−ln(t−1)∣_2 ^(e+1) =∣ln(t^2 /(t−1))∣_2 ^(e+1) =ln(((e+1)^2 )/(e+1−1)) −ln(2^2 /(2−1)) =2ln(e+1)−lne−2ln2 =
∫01ex−1ex+1dxt=ex+1dt=exdxdtt−1=dx∫2e+1t−2t.dtt−1∫2e+12(t−1)−tt(t−1)dt∫2e+12dtt−∫2e+1dtt−12lnt−ln(t−1)∣2e+1=∣lnt2t−1∣2e+1=ln(e+1)2e+1−1−ln222−1=2ln(e+1)−lne−2ln2=

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