Integral of -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 $- du$:

      $$\int u\, du$$

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

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

         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:

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

   Method #2

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

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

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

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

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

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

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

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

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

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

               $$\int u\, du$$

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

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

               Now substitute $u$ back in:

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

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

         Now substitute $u$ back in:

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

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is: