Integral of $$$\frac{1}{9 x^{2} - 4}$$$

Find $$$\int \frac{1}{9 x^{2} - 4}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{9 x^{2} - 4} d x}}} = {\color{red}{\int{\left(- \frac{1}{4 \left(3 x + 2\right)} + \frac{1}{4 \left(3 x - 2\right)}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(- \frac{1}{4 \left(3 x + 2\right)} + \frac{1}{4 \left(3 x - 2\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \int{\frac{1}{4 \left(3 x + 2\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{3 x + 2}$$$:

$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - {\color{red}{\int{\frac{1}{4 \left(3 x + 2\right)} d x}}} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{3 x + 2} d x}}{4}\right)}}$$

Let $$$u=3 x + 2$$$.

Then $$$du=\left(3 x + 2\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Therefore,

$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 x + 2} d x}}}}{4} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$

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

$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$

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

$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$

Recall that $$$u=3 x + 2$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x - 2\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(3 x + 2\right)}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x - 2\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$:

$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + {\color{red}{\int{\frac{1}{4 \left(3 x - 2\right)} d x}}} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + {\color{red}{\left(\frac{\int{\frac{1}{3 x - 2} d x}}{4}\right)}}$$

Let $$$u=3 x - 2$$$.

Then $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Thus,

$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{4} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$

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

$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$

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

$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$

Recall that $$$u=3 x - 2$$$:

$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{12}$$

Therefore,

$$\int{\frac{1}{9 x^{2} - 4} d x} = \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12}$$

Add the constant of integration:

$$\int{\frac{1}{9 x^{2} - 4} d x} = \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12}+C$$

$$$\int \frac{1}{9 x^{2} - 4}\, dx = \left(\frac{\ln\left(\left|{3 x - 2}\right|\right)}{12} - \frac{\ln\left(\left|{3 x + 2}\right|\right)}{12}\right) + C$$$A