Integral of (-log(x))/x^2 dx

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

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

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

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

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

      1. There are multiple ways to do this integral.

         Method #1

         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:

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

         Method #2

         1. Use integration by parts:

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

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

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

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

            1. Let $u = - u$.

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

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

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $- e^{u}$

               Now substitute $u$ back in:

               $$- e^{- u}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = - u$.

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

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

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $- e^{u}$

               Now substitute $u$ back in:

               $$- e^{- u}$$

            So, the result is: $e^{- u}$

      So, the result is: $u e^{- u} + e^{- u}$

   Now substitute $u$ back in:

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

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:

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