Integral of $$$\frac{16}{x^{2} - 16}$$$

Find $$$\int \frac{16}{x^{2} - 16}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=16$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2} - 16}$$$:

$${\color{red}{\int{\frac{16}{x^{2} - 16} d x}}} = {\color{red}{\left(16 \int{\frac{1}{x^{2} - 16} d x}\right)}}$$

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

$$16 {\color{red}{\int{\frac{1}{x^{2} - 16} d x}}} = 16 {\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}}$$

Integrate term by term:

$$16 {\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}} = 16 {\color{red}{\left(\int{\frac{1}{8 \left(x - 4\right)} d x} - \int{\frac{1}{8 \left(x + 4\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 4}$$$:

$$16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 16 {\color{red}{\int{\frac{1}{8 \left(x + 4\right)} d x}}} = 16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 16 {\color{red}{\left(\frac{\int{\frac{1}{x + 4} d x}}{8}\right)}}$$

Let $$$u=x + 4$$$.

Then $$$du=\left(x + 4\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$$16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 2 {\color{red}{\int{\frac{1}{x + 4} d x}}} = 16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}} = 16 \int{\frac{1}{8 \left(x - 4\right)} d x} - 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=x + 4$$$:

$$- 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + 16 \int{\frac{1}{8 \left(x - 4\right)} d x} = - 2 \ln{\left(\left|{{\color{red}{\left(x + 4\right)}}}\right| \right)} + 16 \int{\frac{1}{8 \left(x - 4\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 4}$$$:

$$- 2 \ln{\left(\left|{x + 4}\right| \right)} + 16 {\color{red}{\int{\frac{1}{8 \left(x - 4\right)} d x}}} = - 2 \ln{\left(\left|{x + 4}\right| \right)} + 16 {\color{red}{\left(\frac{\int{\frac{1}{x - 4} d x}}{8}\right)}}$$

Let $$$u=x - 4$$$.

Then $$$du=\left(x - 4\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

So,

$$- 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 {\color{red}{\int{\frac{1}{x - 4} d x}}} = - 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$- 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 {\color{red}{\int{\frac{1}{u} d u}}} = - 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=x - 4$$$:

$$- 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - 2 \ln{\left(\left|{x + 4}\right| \right)} + 2 \ln{\left(\left|{{\color{red}{\left(x - 4\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{16}{x^{2} - 16} d x} = 2 \ln{\left(\left|{x - 4}\right| \right)} - 2 \ln{\left(\left|{x + 4}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{16}{x^{2} - 16} d x} = 2 \ln{\left(\left|{x - 4}\right| \right)} - 2 \ln{\left(\left|{x + 4}\right| \right)}+C$$

$$$\int \frac{16}{x^{2} - 16}\, dx = \left(2 \ln\left(\left|{x - 4}\right|\right) - 2 \ln\left(\left|{x + 4}\right|\right)\right) + C$$$A