Integral of dx/(5-3*x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 5 - 3 x$.

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

      $$\int \left(- \frac{1}{3 u}\right)\, du$$

      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:

      $$- \frac{\log{\left(5 - 3 x \right)}}{3}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{5 - 3 x} = - \frac{1}{3 x - 5}$$

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

      $$\int \left(- \frac{1}{3 x - 5}\right)\, dx = - \int \frac{1}{3 x - 5}\, dx$$

      1. Let $u = 3 x - 5$.

         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:

         $$\frac{\log{\left(3 x - 5 \right)}}{3}$$

      So, the result is: $- \frac{\log{\left(3 x - 5 \right)}}{3}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{1}{5 - 3 x} = - \frac{1}{3 x - 5}$$

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

      $$\int \left(- \frac{1}{3 x - 5}\right)\, dx = - \int \frac{1}{3 x - 5}\, dx$$

      1. Let $u = 3 x - 5$.

         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:

         $$\frac{\log{\left(3 x - 5 \right)}}{3}$$

      So, the result is: $- \frac{\log{\left(3 x - 5 \right)}}{3}$

2. Add the constant of integration:

   $$- \frac{\log{\left(5 - 3 x \right)}}{3}+ \mathrm{constant}$$

The answer is:

$$- \frac{\log{\left(5 - 3 x \right)}}{3}+ \mathrm{constant}$$