Integral of 1/x(√x-1/x+1) 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 $du$:

      $$\int \frac{u - \sqrt{\frac{1}{u}} - 1}{u}\, du$$

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

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

         $$\int \frac{u^{\frac{3}{2}} + u - 1}{u^{2}}\, du$$

         1. Rewrite the integrand:

            $$\frac{u^{\frac{3}{2}} + u - 1}{u^{2}} = \frac{1}{u} - \frac{1}{u^{2}} + \frac{1}{\sqrt{u}}$$

         2. Integrate term-by-term:

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

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

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

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

                  $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

               So, the result is: $\frac{1}{u}$

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

               $$\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$$

            The result is: $2 \sqrt{u} + \log{\left(u \right)} + \frac{1}{u}$

         Now substitute $u$ back in:

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

      Now substitute $u$ back in:

      $$2 \sqrt{x} + \log{\left(x \right)} + \frac{1}{x}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{\left(\sqrt{x} - \frac{1}{x}\right) + 1}{x} = \frac{x^{\frac{3}{2}} + x - 1}{x^{2}}$$

   2. Rewrite the integrand:

      $$\frac{x^{\frac{3}{2}} + x - 1}{x^{2}} = \frac{1}{x} - \frac{1}{x^{2}} + \frac{1}{\sqrt{x}}$$

   3. Integrate term-by-term:

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

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

         $$\int \left(- \frac{1}{x^{2}}\right)\, dx = - \int \frac{1}{x^{2}}\, dx$$

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

            $$\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}$$

         So, the result is: $\frac{1}{x}$

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

         $$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x}$$

      The result is: $2 \sqrt{x} + \log{\left(x \right)} + \frac{1}{x}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{\left(\sqrt{x} - \frac{1}{x}\right) + 1}{x} = \frac{1}{x} - \frac{1}{x^{2}} + \frac{1}{\sqrt{x}}$$

   2. Integrate term-by-term:

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

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

         $$\int \left(- \frac{1}{x^{2}}\right)\, dx = - \int \frac{1}{x^{2}}\, dx$$

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

            $$\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}$$

         So, the result is: $\frac{1}{x}$

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

         $$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x}$$

      The result is: $2 \sqrt{x} + \log{\left(x \right)} + \frac{1}{x}$

2. Add the constant of integration:

   $$2 \sqrt{x} + \log{\left(x \right)} + \frac{1}{x}+ \mathrm{constant}$$

The answer is:

$$2 \sqrt{x} + \log{\left(x \right)} + \frac{1}{x}+ \mathrm{constant}$$