Integral of (x^2-2x+3)ln(x) 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 $du$:

      $$\int \left(u e^{3 u} - 2 u e^{2 u} + 3 u e^{u}\right)\, du$$

      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

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

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

            1. There are multiple ways to do this integral.

               Method #1

               1. Let $u = 3 u$.

                  Then let $du = 3 du$ and substitute $\frac{du}{3}$:

                  $$\int \frac{e^{u}}{9}\, du$$

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

                     $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                     1. The integral of the exponential function is itself.

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

                     So, the result is: $\frac{e^{u}}{3}$

                  Now substitute $u$ back in:

                  $$\frac{e^{3 u}}{3}$$

               Method #2

               1. Let $u = e^{3 u}$.

                  Then let $du = 3 e^{3 u} du$ and substitute $\frac{du}{3}$:

                  $$\int \frac{1}{9}\, du$$

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

                     $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

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

                        $$\int 1\, du = u$$

                     So, the result is: $\frac{u}{3}$

                  Now substitute $u$ back in:

                  $$\frac{e^{3 u}}{3}$$

            Now evaluate the sub-integral.

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

            $$\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$$

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

                  $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            So, the result is: $\frac{e^{3 u}}{9}$

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

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

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

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               Now evaluate the sub-integral.

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               So, the result is: $\frac{e^{2 u}}{4}$

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

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

            $$\int 3 u e^{u}\, du = 3 \int u e^{u}\, du$$

            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: $3 u e^{u} - 3 e^{u}$

         The result is: $\frac{u e^{3 u}}{3} - u e^{2 u} + 3 u e^{u} - \frac{e^{3 u}}{9} + \frac{e^{2 u}}{2} - 3 e^{u}$

      Now substitute $u$ back in:

      $$\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 3 x \log{\left(x \right)} - 3 x$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         $$\int u e^{3 u}\, du$$

         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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

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

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

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

                  $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            Now evaluate the sub-integral.

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

            $$\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$$

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

                  $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            So, the result is: $\frac{e^{3 u}}{9}$

         Now substitute $u$ back in:

         $$\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$$

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

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

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

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

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

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

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               Now evaluate the sub-integral.

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               So, the result is: $\frac{e^{2 u}}{4}$

            Now substitute $u$ back in:

            $$\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$$

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

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

         $$\int 3 \log{\left(x \right)}\, dx = 3 \int \log{\left(x \right)}\, dx$$

         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: $3 x \log{\left(x \right)} - 3 x$

      The result is: $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 3 x \log{\left(x \right)} - 3 x$

   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 $\operatorname{dv}{\left(x \right)} = x^{2} - 2 x + 3$.

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

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

      1. Integrate term-by-term:

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

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

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

            $$\int \left(- 2 x\right)\, dx = - 2 \int x\, dx$$

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

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

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

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

            $$\int 3\, dx = 3 x$$

         The result is: $\frac{x^{3}}{3} - x^{2} + 3 x$

      Now evaluate the sub-integral.

   2. Rewrite the integrand:

      $$\frac{\frac{x^{3}}{3} - x^{2} + 3 x}{x} = \frac{x^{2}}{3} - x + 3$$

   3. Integrate term-by-term:

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

         $$\int \frac{x^{2}}{3}\, dx = \frac{\int x^{2}\, dx}{3}$$

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

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         So, the result is: $\frac{x^{3}}{9}$

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

         $$\int \left(- x\right)\, dx = - \int x\, dx$$

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

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

         So, the result is: $- \frac{x^{2}}{2}$

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

         $$\int 3\, dx = 3 x$$

      The result is: $\frac{x^{3}}{9} - \frac{x^{2}}{2} + 3 x$

   Method #4

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         $$\int u e^{3 u}\, du$$

         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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

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

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

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

                  $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            Now evaluate the sub-integral.

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

            $$\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$$

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

                  $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            So, the result is: $\frac{e^{3 u}}{9}$

         Now substitute $u$ back in:

         $$\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$$

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

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

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

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

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

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

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               Now evaluate the sub-integral.

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

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

               1. Let $u = 2 u$.

                  Then let $du = 2 du$ and substitute $\frac{du}{2}$:

                  $$\int \frac{e^{u}}{4}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

                  $$\frac{e^{2 u}}{2}$$

               So, the result is: $\frac{e^{2 u}}{4}$

            Now substitute $u$ back in:

            $$\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$$

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

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

         $$\int 3 \log{\left(x \right)}\, dx = 3 \int \log{\left(x \right)}\, dx$$

         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: $3 x \log{\left(x \right)} - 3 x$

      The result is: $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 3 x \log{\left(x \right)} - 3 x$

2. Now simplify:

   $$\frac{x \left(6 x^{2} \log{\left(x \right)} - 2 x^{2} - 18 x \log{\left(x \right)} + 9 x + 54 \log{\left(x \right)} - 54\right)}{18}$$

3. Add the constant of integration:

   $$\frac{x \left(6 x^{2} \log{\left(x \right)} - 2 x^{2} - 18 x \log{\left(x \right)} + 9 x + 54 \log{\left(x \right)} - 54\right)}{18}+ \mathrm{constant}$$

The answer is: