Integral of ln(x)/(x^2) 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 $du$:

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

      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:

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

   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 $\operatorname{dv}{\left(x \right)} = \frac{1}{x^{2}}$.

      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 $n \neq -1$:

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

      Now evaluate the sub-integral.

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

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

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

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

      So, the result is: $\frac{1}{x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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