Integral of (2+sqrt(lnx))/x dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \frac{dx}{2 x \sqrt{\log{\left(x \right)}}}$ and substitute $du$:

      $$\int \left(2 u^{2} + 4 u\right)\, du$$

      1. Integrate term-by-term:

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

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

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

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            So, the result is: $\frac{2 u^{3}}{3}$

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

            $$\int 4 u\, du = 4 \int u\, du$$

            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: $2 u^{2}$

         The result is: $\frac{2 u^{3}}{3} + 2 u^{2}$

      Now substitute $u$ back in:

      $$\frac{2 \log{\left(x \right)}^{\frac{3}{2}}}{3} + 2 \log{\left(x \right)}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

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

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

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

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

                  $$\int \sqrt{u}\, du = - \int \sqrt{u}\, du$$

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

                     $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

                  So, the result is: $- \frac{2 u^{\frac{3}{2}}}{3}$

               Now substitute $u$ back in:

               $$- \frac{2 \log{\left(\frac{1}{u} \right)}^{\frac{3}{2}}}{3}$$

            So, the result is: $\frac{2 \log{\left(\frac{1}{u} \right)}^{\frac{3}{2}}}{3}$

         Now substitute $u$ back in:

         $$\frac{2 \log{\left(x \right)}^{\frac{3}{2}}}{3}$$

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

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

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

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

      The result is: $\frac{2 \log{\left(x \right)}^{\frac{3}{2}}}{3} + 2 \log{\left(x \right)}$

2. Add the constant of integration:

   $$\frac{2 \log{\left(x \right)}^{\frac{3}{2}}}{3} + 2 \log{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$\frac{2 \log{\left(x \right)}^{\frac{3}{2}}}{3} + 2 \log{\left(x \right)}+ \mathrm{constant}$$