Integral of ln(1/x) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

         1. There are multiple ways to do this integral.

            Method #1

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

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

               $$\int u e^{- u}\, du$$

               1. Let $u = - u$.

                  Then let $du = - du$ and substitute $du$:

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

                  1. Use integration by parts:

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

                     Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

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

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

                     1. The integral of the exponential function is itself.

                        $$\int e^{u}\, du = e^{u}$$

                     Now evaluate the sub-integral.

                  2. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  Now substitute $u$ back in:

                  $$- u e^{- u} - e^{- u}$$

               Now substitute $u$ back in:

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

            Method #2

            1. Use integration by parts:

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

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

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

               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:

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

               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: $\frac{1}{u}$

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

      Now substitute $u$ back in:

      $$- x \log{\left(x \right)} + x$$

   Method #2

   1. Use integration by parts:

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

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

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

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

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

   2. The integral of a constant is the constant times the variable of integration:

      $$\int \left(-1\right)\, dx = - x$$

2. Now simplify:

   $$x \left(1 - \log{\left(x \right)}\right)$$

3. Add the constant of integration:

   $$x \left(1 - \log{\left(x \right)}\right)+ \mathrm{constant}$$

The answer is:

$$x \left(1 - \log{\left(x \right)}\right)+ \mathrm{constant}$$