∫1/1+e^x dx = ?

 

Asked by anannyaratha.2004 | 19th Jul, 2021, 01:36: PM

Expert Answer:

begin mathsize 14px style I space equals space integral fraction numerator d x over denominator 1 plus e to the power of x end fraction space equals space integral fraction numerator e to the power of x d x over denominator space e to the power of x left parenthesis 1 plus e to the power of x right parenthesis end fraction end style
 
Let u = ex  , then du = ex dx 
 
begin mathsize 14px style I space equals space integral fraction numerator d u over denominator u space left parenthesis 1 plus u right parenthesis end fraction space equals space integral fraction numerator d u over denominator u end fraction space minus space integral fraction numerator d u over denominator 1 plus u end fraction end style
begin mathsize 14px style I space equals space log space u space minus space log left parenthesis 1 plus u right parenthesis space equals space log open parentheses fraction numerator e to the power of x over denominator e to the power of x plus 1 end fraction close parentheses space plus space C end style

Answered by Thiyagarajan K | 17th Sep, 2021, 03:46: PM

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