\int \frac{x+1-1}{x+1} \, \mathrm{d}x= \int \frac{x+1}{x+1} \, \mathrm{d}x - \int \frac{1}{x+1} \, \mathrm{d}x = \int \mathrm{d}x - \int \frac{ \mathrm{d}x}{x+1} = x - \log(x+1) + C