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the integral of ∫ e^r/1+e^r?

 May 1, 2020
 #1
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 ∫  er  / [ 1 + er]   dr  =   Ln [ 1 + e] + C

 

The key  here is to recognize  that  the derivative of 1 + er   =   er

 

So  remember that  if  u = 1 + er  then

 

 ∫   u '  /  u     du   =   Ln u + C

 

cool cool cool

 May 1, 2020

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