Integral of y=2log(2x+1) dx

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

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

   1. There are multiple ways to do this integral.

      Method #1

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

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

         $$\int \frac{\log{\left(u \right)}}{2}\, du$$

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

            $$\int \log{\left(u \right)}\, du = \frac{\int \log{\left(u \right)}\, du}{2}$$

            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: $\frac{u \log{\left(u \right)}}{2} - \frac{u}{2}$

         Now substitute $u$ back in:

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

      Method #2

      1. Use integration by parts:

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

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

         Then $\operatorname{du}{\left(x \right)} = \frac{2}{2 x + 1}$.

         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:

         $$\int \frac{2 x}{2 x + 1}\, dx = 2 \int \frac{x}{2 x + 1}\, dx$$

         1. Rewrite the integrand:

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

         2. Integrate term-by-term:

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

               $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

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

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

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

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

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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