Integral of xlog(x)-x+c1x+c2 dx

1. Integrate term-by-term:

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

      $$\int c_{2}\, dx = c_{2} x$$

   1. Integrate term-by-term:

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

         $$\int c_{1} x\, dx = c_{1} \int x\, dx$$

         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: $\frac{c_{1} x^{2}}{2}$

      1. Integrate term-by-term:

         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 $du$:

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

               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 $\frac{du}{2}$:

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

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

                     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 $\frac{du}{2}$:

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

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

                     Now substitute $u$ back in:

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

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

               Now substitute $u$ back in:

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

            Method #2

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

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

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

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

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

               Now evaluate the sub-integral.

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

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

               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: $\frac{x^{2}}{4}$

         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: $- \frac{x^{2}}{2}$

         The result is: $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$

      The result is: $\frac{c_{1} x^{2}}{2} + \frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$

   The result is: $\frac{c_{1} x^{2}}{2} + c_{2} x + \frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$

2. Now simplify:

   $$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}+ \mathrm{constant}$$