Integral of log(x)^9/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 $du$:

      $$\int u^{9}\, du$$

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

         $$\int u^{9}\, du = \frac{u^{10}}{10}$$

      Now substitute $u$ back in:

      $$\frac{\log{\left(x \right)}^{10}}{10}$$

   Method #2

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

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

      $$\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{9}}{u}\right)\, du$$

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

         $$\int \frac{\log{\left(\frac{1}{u} \right)}^{9}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{9}}{u}\, du$$

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

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

            $$\int \left(- u^{9}\right)\, du$$

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

               $$\int u^{9}\, du = - \int u^{9}\, du$$

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

                  $$\int u^{9}\, du = \frac{u^{10}}{10}$$

               So, the result is: $- \frac{u^{10}}{10}$

            Now substitute $u$ back in:

            $$- \frac{\log{\left(\frac{1}{u} \right)}^{10}}{10}$$

         So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{10}}{10}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(x \right)}^{10}}{10}$$

2. Add the constant of integration:

   $$\frac{\log{\left(x \right)}^{10}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(x \right)}^{10}}{10}+ \mathrm{constant}$$