Integral of 1/x(xln(x)-x+c) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

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

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

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

            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 is the constant times the variable of integration:

                  $$\int \left(- c\right)\, du = - u c$$

               1. The integral of the exponential function is itself.

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

               The result is: $- u c - u e^{u} + 2 e^{u}$

            Now substitute $u$ back in:

            $$- c \log{\left(\frac{1}{u} \right)} - \frac{\log{\left(\frac{1}{u} \right)}}{u} + \frac{2}{u}$$

         So, the result is: $c \log{\left(\frac{1}{u} \right)} + \frac{\log{\left(\frac{1}{u} \right)}}{u} - \frac{2}{u}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

         $$\int \frac{c}{x}\, dx = c \int \frac{1}{x}\, dx$$

         1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$.

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

      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:

         $$\int \left(-1\right)\, dx = - x$$

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

2. Add the constant of integration:

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

The answer is: