Integral of x^(-0,5)*ln(x) dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = - \frac{dx}{2 x^{\frac{3}{2}}}$ and substitute $- 2 du$:

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

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

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

         1. Use integration by parts:

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

            Let $u{\left(u \right)} = \log{\left(\frac{1}{u^{2}} \right)}$ and let $\operatorname{dv}{\left(u \right)} = \frac{1}{u^{2}}$.

            Then $\operatorname{du}{\left(u \right)} = - \frac{2}{u}$.

            To find $v{\left(u \right)}$:

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

               $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

            Now evaluate the sub-integral.

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

            $$\int \frac{2}{u^{2}}\, du = 2 \int \frac{1}{u^{2}}\, du$$

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

               $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

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

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

      Now substitute $u$ back in:

      $$2 \sqrt{x} \log{\left(x \right)} - 4 \sqrt{x}$$

   Method #2

   1. Use integration by parts:

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

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

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

      To find $v{\left(x \right)}$:

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

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

      Now evaluate the sub-integral.

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

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

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

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

      So, the result is: $4 \sqrt{x}$

2. Now simplify:

   $$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)$$

3. Add the constant of integration:

   $$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)+ \mathrm{constant}$$

The answer is:

$$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)+ \mathrm{constant}$$