Integral of sqrt(lnx)/x dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

      Now substitute $u$ back in:

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

   Method #2

   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:

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

2. Add the constant of integration:

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

The answer is:

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