Integral of ln(4x+2) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 4 x + 2$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{\log{\left(u \right)}}{4}\, du$$

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

         $$\int \log{\left(u \right)}\, du = \frac{\int \log{\left(u \right)}\, du}{4}$$

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$.

            Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$.

            To find $v{\left(u \right)}$:

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            Now evaluate the sub-integral.

         2. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, du = u$$

         So, the result is: $\frac{u \log{\left(u \right)}}{4} - \frac{u}{4}$

      Now substitute $u$ back in:

      $$- x + \frac{\left(4 x + 2\right) \log{\left(4 x + 2 \right)}}{4} - \frac{1}{2}$$

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(4 x + 2 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{4}{4 x + 2}$.

      To find $v{\left(x \right)}$:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

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

      $$\int \frac{4 x}{4 x + 2}\, dx = 4 \int \frac{x}{4 x + 2}\, dx$$

      1. Rewrite the integrand:

         $$\frac{x}{4 x + 2} = \frac{1}{4} - \frac{1}{4 \left(2 x + 1\right)}$$

      2. Integrate term-by-term:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int \frac{1}{4}\, dx = \frac{x}{4}$$

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

            $$\int \left(- \frac{1}{4 \left(2 x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{2 x + 1}\, dx}{4}$$

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

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{1}{2 u}\, du$$

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

                  $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

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

                  So, the result is: $\frac{\log{\left(u \right)}}{2}$

               Now substitute $u$ back in:

               $$\frac{\log{\left(2 x + 1 \right)}}{2}$$

            So, the result is: $- \frac{\log{\left(2 x + 1 \right)}}{8}$

         The result is: $\frac{x}{4} - \frac{\log{\left(2 x + 1 \right)}}{8}$

      So, the result is: $x - \frac{\log{\left(2 x + 1 \right)}}{2}$

2. Now simplify:

   $$- x + \frac{\left(2 x + 1\right) \log{\left(4 x + 2 \right)}}{2} - \frac{1}{2}$$

3. Add the constant of integration:

   $$- x + \frac{\left(2 x + 1\right) \log{\left(4 x + 2 \right)}}{2} - \frac{1}{2}+ \mathrm{constant}$$

The answer is:

$$- x + \frac{\left(2 x + 1\right) \log{\left(4 x + 2 \right)}}{2} - \frac{1}{2}+ \mathrm{constant}$$