Integral of (log(x)-1)/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 $du$:

      $$\int \left(u - 1\right) e^{- u}\, du$$

      1. Let $u = - u$.

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

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

         1. Integrate term-by-term:

            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.

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

            1. The integral of the exponential function is itself.

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

            The result is: $u e^{u}$

         Now substitute $u$ back in:

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

      $$\frac{\log{\left(x \right)} - 1}{x^{2}} = \frac{\log{\left(x \right)}}{x^{2}} - \frac{1}{x^{2}}$$

   2. Integrate term-by-term:

      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:

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

      1. 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: $\frac{1}{x}$

      The result is: $- \frac{\log{\left(x \right)}}{x}$

2. Add the constant of integration:

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

The answer is:

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