Integral of lnx-(lnx)^2 dx

1. Integrate term-by-term:

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

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

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

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

         $$\int u^{2} e^{u}\, du$$

         1. Use integration by parts:

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

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

            Then $\operatorname{du}{\left(u \right)} = 2 u$.

            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. Use integration by parts:

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

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

            Then $\operatorname{du}{\left(u \right)} = 2$.

            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.

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

            $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $2 e^{u}$

         Now substitute $u$ back in:

         $$x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$$

      So, the result is: $- x \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)} - 2 x$

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

      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:

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

   The result is: $- x \log{\left(x \right)}^{2} + 3 x \log{\left(x \right)} - 3 x$

2. Now simplify:

   $$x \left(- \log{\left(x \right)}^{2} + 3 \log{\left(x \right)} - 3\right)$$

3. Add the constant of integration:

   $$x \left(- \log{\left(x \right)}^{2} + 3 \log{\left(x \right)} - 3\right)+ \mathrm{constant}$$

The answer is:

$$x \left(- \log{\left(x \right)}^{2} + 3 \log{\left(x \right)} - 3\right)+ \mathrm{constant}$$