Integral of $$$\frac{1}{a + x}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{a + x}\, dx$$$.

Let $$$u=a + x$$$.

Then $$$du=\left(a + x\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

So,

$${\color{red}{\int{\frac{1}{a + x} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=a + x$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(a + x\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{1}{a + x} d x} = \ln{\left(\left|{a + x}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{1}{a + x} d x} = \ln{\left(\left|{a + x}\right| \right)}+C$$

$$$\int \frac{1}{a + x}\, dx = \ln\left(\left|{a + x}\right|\right) + C$$$A