Integral of (x^2-9)/(x+3) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\frac{x^{2} - 9}{x + 3} = x - 3$$

   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:

         $$\int \left(-3\right)\, dx = - 3 x$$

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

   Method #2

   1. Rewrite the integrand:

      $$\frac{x^{2} - 9}{x + 3} = \frac{x^{2}}{x + 3} - \frac{9}{x + 3}$$

   2. Integrate term-by-term:

      1. Rewrite the integrand:

         $$\frac{x^{2}}{x + 3} = x - 3 + \frac{9}{x + 3}$$

      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:

            $$\int \left(-3\right)\, dx = - 3 x$$

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

            $$\int \frac{9}{x + 3}\, dx = 9 \int \frac{1}{x + 3}\, dx$$

            1. Let $u = x + 3$.

               Then let $du = dx$ and substitute $du$:

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

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

               Now substitute $u$ back in:

               $$\log{\left(x + 3 \right)}$$

            So, the result is: $9 \log{\left(x + 3 \right)}$

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

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

         $$\int \left(- \frac{9}{x + 3}\right)\, dx = - 9 \int \frac{1}{x + 3}\, dx$$

         1. Let $u = x + 3$.

            Then let $du = dx$ and substitute $du$:

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

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

            Now substitute $u$ back in:

            $$\log{\left(x + 3 \right)}$$

         So, the result is: $- 9 \log{\left(x + 3 \right)}$

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

2. Now simplify:

   $$\frac{x \left(x - 6\right)}{2}$$

3. Add the constant of integration:

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

The answer is:

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