Integral of ((x-3)^2/x)*dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\left(x - 3\right)^{2} \cdot \frac{1}{x} 1 = x - 6 + \frac{9}{x}$$

   2. Integrate term-by-term:

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

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

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

         $$\int \left(-6\right)\, dx = - 6 x$$

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

         $$\int \frac{9}{x}\, dx = 9 \int \frac{1}{x}\, dx$$

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

         So, the result is: $9 \log{\left(x \right)}$

      The result is: $\frac{x^{2}}{2} - 6 x + 9 \log{\left(x \right)}$

   Method #2

   1. Rewrite the integrand:

      $$\left(x - 3\right)^{2} \cdot \frac{1}{x} 1 = \frac{x^{2} - 6 x + 9}{x}$$

   2. Rewrite the integrand:

      $$\frac{x^{2} - 6 x + 9}{x} = x - 6 + \frac{9}{x}$$

   3. Integrate term-by-term:

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

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

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

         $$\int \left(-6\right)\, dx = - 6 x$$

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

         $$\int \frac{9}{x}\, dx = 9 \int \frac{1}{x}\, dx$$

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

         So, the result is: $9 \log{\left(x \right)}$

      The result is: $\frac{x^{2}}{2} - 6 x + 9 \log{\left(x \right)}$

2. Add the constant of integration:

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

The answer is:

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