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Question Number 84702 by Power last updated on 15/Mar/20

Commented by abdomathmax last updated on 15/Mar/20

I =∫   (dx/(shx+1)) ⇒I=∫  (dx/(((e^x −e^(−x) )/2)+1)) =∫((2dx)/(e^x −e^(−x)  +2))  =_(e^x =t)      ∫  (2/(t−t^(−1)  +2))(dt/t) =∫ ((2dt)/(t^2 −1+2t))  =∫ ((2dt)/(t^2  +2t−1)) =∫ ((2dt)/((t+1)^2 −2)) =∫ ((2dt)/((t+1−(√2))(t+1+(√2))))  =(2/(2(√2)))∫  ((1/(t+1−(√2)))−(1/(t+1+(√2))))dt  =(1/(√2))ln∣((t+1−(√2))/(t+1+(√2)))∣ +C  =(1/(√2))ln∣((e^x  +1−(√2))/(e^x  +1+(√2)))∣ +C

I=dxshx+1I=dxexex2+1=2dxexex+2=ex=t2tt1+2dtt=2dtt21+2t=2dtt2+2t1=2dt(t+1)22=2dt(t+12)(t+1+2)=222(1t+121t+1+2)dt=12lnt+12t+1+2+C=12lnex+12ex+1+2+C

Commented by Power last updated on 15/Mar/20

thanks

thanks

Commented by abdomathmax last updated on 15/Mar/20

you are welcome

youarewelcome

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