Integral of (sqrt(2+4*ln(x)))/x dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \frac{\sqrt{u}}{4}\, du$$

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

         $$\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{4}$$

         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: $\frac{u^{\frac{3}{2}}}{6}$

      Now substitute $u$ back in:

      $$\frac{\left(4 \log{\left(x \right)} + 2\right)^{\frac{3}{2}}}{6}$$

   Method #2

   1. Rewrite the integrand:

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

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

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

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

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

         $$\int \left(- \frac{\sqrt{2 \log{\left(\frac{1}{u} \right)} + 1}}{u}\right)\, du$$

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

            $$\int \frac{\sqrt{2 \log{\left(\frac{1}{u} \right)} + 1}}{u}\, du = - \int \frac{\sqrt{2 \log{\left(\frac{1}{u} \right)} + 1}}{u}\, du$$

            1. Let $u = 2 \log{\left(\frac{1}{u} \right)} + 1$.

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

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

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

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

                  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: $- \frac{u^{\frac{3}{2}}}{3}$

               Now substitute $u$ back in:

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

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

         Now substitute $u$ back in:

         $$\frac{\left(2 \log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}$$

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

2. Now simplify:

   $$\frac{\sqrt{2} \left(2 \log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}$$

3. Add the constant of integration:

   $$\frac{\sqrt{2} \left(2 \log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{\sqrt{2} \left(2 \log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$