Integral of 1/(18*x+2) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 18 x + 2$.

      Then let $du = 18 dx$ and substitute $\frac{du}{18}$:

      $$\int \frac{1}{18 u}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{18}$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         So, the result is: $\frac{\log{\left(u \right)}}{18}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(18 x + 2 \right)}}{18}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{18 x + 2} = \frac{1}{2 \left(9 x + 1\right)}$$

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{1}{2 \left(9 x + 1\right)}\, dx = \frac{\int \frac{1}{9 x + 1}\, dx}{2}$$

      1. Let $u = 9 x + 1$.

         Then let $du = 9 dx$ and substitute $\frac{du}{9}$:

         $$\int \frac{1}{9 u}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{9}$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $\frac{\log{\left(u \right)}}{9}$

         Now substitute $u$ back in:

         $$\frac{\log{\left(9 x + 1 \right)}}{9}$$

      So, the result is: $\frac{\log{\left(9 x + 1 \right)}}{18}$

2. Now simplify:

   $$\frac{\log{\left(18 x + 2 \right)}}{18}$$

3. Add the constant of integration:

   $$\frac{\log{\left(18 x + 2 \right)}}{18}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(18 x + 2 \right)}}{18}+ \mathrm{constant}$$