1.

Integrate from 0 to 1 (x/1+x)\(\int\limits_0^1\frac{x}{1+x}dx\)

Answer»

\(\int\limits_0^1\frac{x}{1+x}dx\)

 = \(\int\limits_0^1\frac{1+x-1}{1+x}dx\) 

 =  \(\int\limits_0^1(1+\frac{1}{1+x})dx\)

 = \( [x - log(1 + x)]_0^1\) 

 = (1 - 0) -(log 2 - log 1)

 = 1 - log 2 = log e - log 2

 = log(e/2)



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