Integral of 1/x(ln(x)+4) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \frac{1}{x}$.

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

      $$\int \left(- \frac{\log{\left(\frac{1}{u} \right)} + 4}{u}\right)\, du$$

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

         $$\int \frac{\log{\left(\frac{1}{u} \right)} + 4}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)} + 4}{u}\, du$$

         1. Let $u = \log{\left(\frac{1}{u} \right)} + 4$.

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

            $$\int \left(- u\right)\, du$$

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

               $$\int u\, du = - \int u\, du$$

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

               So, the result is: $- \frac{u^{2}}{2}$

            Now substitute $u$ back in:

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

         So, the result is: $\frac{\left(\log{\left(\frac{1}{u} \right)} + 4\right)^{2}}{2}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

      $$\frac{\log{\left(x \right)} + 4}{x} = \frac{\log{\left(x \right)}}{x} + \frac{4}{x}$$

   2. Integrate term-by-term:

      1. Let $u = \frac{1}{x}$.

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

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

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

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

            1. Let $u = \log{\left(\frac{1}{u} \right)}$.

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

               $$\int \left(- u\right)\, du$$

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

                  $$\int u\, du = - \int u\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

                  So, the result is: $- \frac{u^{2}}{2}$

               Now substitute $u$ back in:

               $$- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$$

            So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$

         Now substitute $u$ back in:

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

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

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

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

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

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

2. Add the constant of integration:

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

The answer is:

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