Integral of 1/(1000+6x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 6 x + 1000$.

      Then let $du = 6 dx$ and substitute $\frac{du}{6}$:

      $$\int \frac{1}{6 u}\, du$$

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

         $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{6}$$

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

         So, the result is: $\frac{\log{\left(u \right)}}{6}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(6 x + 1000 \right)}}{6}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{6 x + 1000} = \frac{1}{2 \left(3 x + 500\right)}$$

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

      $$\int \frac{1}{2 \left(3 x + 500\right)}\, dx = \frac{\int \frac{1}{3 x + 500}\, dx}{2}$$

      1. Let $u = 3 x + 500$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{1}{3 u}\, du$$

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

            $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{3}$$

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

            So, the result is: $\frac{\log{\left(u \right)}}{3}$

         Now substitute $u$ back in:

         $$\frac{\log{\left(3 x + 500 \right)}}{3}$$

      So, the result is: $\frac{\log{\left(3 x + 500 \right)}}{6}$

2. Add the constant of integration:

   $$\frac{\log{\left(6 x + 1000 \right)}}{6}+ \mathrm{constant}$$

The answer is: