# How do you integrate (1+x) /(1+ x)^2 dx?

$\frac{1 + x}{1 + x} ^ 2 = \frac{1}{1 + x}$
To do this, you have to point out that the domain of the function is $D = \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$.
$\int \frac{1 + x}{1 + x} ^ 2 \mathrm{dx} = \int \frac{1}{1 + x} \mathrm{dx} = \ln | 1 + x | + C$