Integral of $$$\frac{1}{x^{2} - 9}$$$

Find $$$\int \frac{1}{x^{2} - 9}\, dx$$$.

Perform partial fraction decomposition (steps can be seen »):

$${\color{red}{\int{\frac{1}{x^{2} - 9} d x}}} = {\color{red}{\int{\left(- \frac{1}{6 \left(x + 3\right)} + \frac{1}{6 \left(x - 3\right)}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(- \frac{1}{6 \left(x + 3\right)} + \frac{1}{6 \left(x - 3\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{6 \left(x - 3\right)} d x} - \int{\frac{1}{6 \left(x + 3\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 3}$$$:

$$\int{\frac{1}{6 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{6 \left(x + 3\right)} d x}}} = \int{\frac{1}{6 \left(x - 3\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x + 3} d x}}{6}\right)}}$$

Let $$$u=x + 3$$$.

Then $$$du=\left(x + 3\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Therefore,

$$\int{\frac{1}{6 \left(x - 3\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 3} d x}}}}{6} = \int{\frac{1}{6 \left(x - 3\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\int{\frac{1}{6 \left(x - 3\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6} = \int{\frac{1}{6 \left(x - 3\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recall that $$$u=x + 3$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} + \int{\frac{1}{6 \left(x - 3\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x + 3\right)}}}\right| \right)}}{6} + \int{\frac{1}{6 \left(x - 3\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:

$$- \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + {\color{red}{\int{\frac{1}{6 \left(x - 3\right)} d x}}} = - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + {\color{red}{\left(\frac{\int{\frac{1}{x - 3} d x}}{6}\right)}}$$

Let $$$u=x - 3$$$.

Then $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$$- \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{{\color{red}{\int{\frac{1}{x - 3} d x}}}}{6} = - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6} = - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recall that $$$u=x - 3$$$:

$$- \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} = - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)}}{6}$$

Therefore,

$$\int{\frac{1}{x^{2} - 9} d x} = \frac{\ln{\left(\left|{x - 3}\right| \right)}}{6} - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6}$$

Add the constant of integration:

$$\int{\frac{1}{x^{2} - 9} d x} = \frac{\ln{\left(\left|{x - 3}\right| \right)}}{6} - \frac{\ln{\left(\left|{x + 3}\right| \right)}}{6}+C$$

$$$\int \frac{1}{x^{2} - 9}\, dx = \left(\frac{\ln\left(\left|{x - 3}\right|\right)}{6} - \frac{\ln\left(\left|{x + 3}\right|\right)}{6}\right) + C$$$A