integrate x/(2+x)

∫[ x/(2 + x) ] dx

Let  u  =  2 + x   --->  x  =  u - 2         and        du  =  dx

∫[ x/(2 + x) ] dx   --->   ∫[ (u - 2)/u ] du   --->   ∫[ u/u - 2/u ]du   --->   ∫1 du  -  2∫1/u du

   --->   u  - 2 ln|u| + C   --->   2 + x - 2 ln| 2 + x | + C

Hi Geno,

why have you used     ln|u|     instead of   ln(u)   ?

The general integral of  1/x  is  ln|x|  because ln() must contain a positive value.

If we know that x is positive, we don't need the absolute value bars.

$$\int\frac{1}{x}dx$$

If x is positive we both agree that the answer is ln(x) +c

 If x is negative the answer should be    -ln(-x)+c    OR     -ln|x|+c       shouldn't it ?