Question

$\int\limits^1_0 { \frac{x}{x+1} } \, dx$

Answer 1

x / ( x + 1 ) = 1    -     1 / ( x +1 ) 
integ = integ.[ 1    -  1 / ( x+1)  ]dx=
= x  - ln( x + 1) 
cu val.   x= 0     si x =1 
= 1 - ln2 + ln1 =
                 =0
= lne -ln2 = ln( e/2)

Answer 2

=$\int\limits^1_0 { \frac{x+1-1}{x+1} } \, dx= \int\limits^1_0 {1} \, dx- \int\limits^1_0 { \frac{1}{x+1} } \, dx$
=$x-ln(x+1)$ in limite de la 0 la 1
=1-ln2