Integral of (lnx+2)*lnx 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 $du$:

      $$\int \left(u^{2} e^{u} + 2 u e^{u}\right)\, du$$

      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 $u{\left(u \right)} = u^{2}$ 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: $2 e^{u}$

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

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

            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.

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

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

         The result is: $u^{2} e^{u}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      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:

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

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

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

         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:

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

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

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

   Method #3

   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 $\operatorname{dv}{\left(x \right)} = \log{\left(x \right)} + 2$.

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

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

      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 $u{\left(x \right)} = \log{\left(x \right)}$ 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:

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

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

            $$\int 2\, dx = 2 x$$

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

      Now evaluate the sub-integral.

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

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

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

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

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

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

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

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

            1. Integrate term-by-term:

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

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

                  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.

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

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

               1. 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. The integral of the exponential function is itself.

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

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

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

            Now substitute $u$ back in:

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

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

      Now substitute $u$ back in:

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

   Method #4

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      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:

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

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

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

         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:

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

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

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

2. Add the constant of integration:

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

The answer is:

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