Integral of log(x+1,7) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x + \frac{17}{10}$.

      Then let $du = dx$ and substitute $du$:

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

      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:

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

      Now substitute $u$ back in:

      $$- x + \left(x + \frac{17}{10}\right) \log{\left(x + \frac{17}{10} \right)} - \frac{17}{10}$$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \log{\left(x + \frac{17}{10} \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + \frac{17}{10}}$.

      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. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\frac{x}{x + \frac{17}{10}} = 1 - \frac{17}{10 x + 17}$$

      2. Integrate term-by-term:

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

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

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

            $$\int \left(- \frac{17}{10 x + 17}\right)\, dx = - 17 \int \frac{1}{10 x + 17}\, dx$$

            1. Let $u = 10 x + 17$.

               Then let $du = 10 dx$ and substitute $\frac{du}{10}$:

               $$\int \frac{1}{10 u}\, du$$

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

                  $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{10}$$

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

                  So, the result is: $\frac{\log{\left(u \right)}}{10}$

               Now substitute $u$ back in:

               $$\frac{\log{\left(10 x + 17 \right)}}{10}$$

            So, the result is: $- \frac{17 \log{\left(10 x + 17 \right)}}{10}$

         The result is: $x - \frac{17 \log{\left(10 x + 17 \right)}}{10}$

      Method #2

      1. Rewrite the integrand:

         $$\frac{x}{x + \frac{17}{10}} = \frac{10 x}{10 x + 17}$$

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

         $$\int \frac{10 x}{10 x + 17}\, dx = 10 \int \frac{x}{10 x + 17}\, dx$$

         1. Rewrite the integrand:

            $$\frac{x}{10 x + 17} = \frac{1}{10} - \frac{17}{10 \left(10 x + 17\right)}$$

         2. Integrate term-by-term:

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

               $$\int \frac{1}{10}\, dx = \frac{x}{10}$$

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

               $$\int \left(- \frac{17}{10 \left(10 x + 17\right)}\right)\, dx = - \frac{17 \int \frac{1}{10 x + 17}\, dx}{10}$$

               1. Let $u = 10 x + 17$.

                  Then let $du = 10 dx$ and substitute $\frac{du}{10}$:

                  $$\int \frac{1}{10 u}\, du$$

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

                     $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{10}$$

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

                     So, the result is: $\frac{\log{\left(u \right)}}{10}$

                  Now substitute $u$ back in:

                  $$\frac{\log{\left(10 x + 17 \right)}}{10}$$

               So, the result is: $- \frac{17 \log{\left(10 x + 17 \right)}}{100}$

            The result is: $\frac{x}{10} - \frac{17 \log{\left(10 x + 17 \right)}}{100}$

         So, the result is: $x - \frac{17 \log{\left(10 x + 17 \right)}}{10}$

2. Now simplify:

   $$- x + \frac{\left(10 x + 17\right) \log{\left(x + \frac{17}{10} \right)}}{10} - \frac{17}{10}$$

3. Add the constant of integration:

   $$- x + \frac{\left(10 x + 17\right) \log{\left(x + \frac{17}{10} \right)}}{10} - \frac{17}{10}+ \mathrm{constant}$$

The answer is:

$$- x + \frac{\left(10 x + 17\right) \log{\left(x + \frac{17}{10} \right)}}{10} - \frac{17}{10}+ \mathrm{constant}$$