Integral of $$$\frac{x^{3}}{x^{2} - 9}$$$

Find $$$\int \frac{x^{3}}{x^{2} - 9}\, dx$$$.

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

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

Integrate term by term:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

Let $$$u=x^{2} - 9$$$.

Then $$$du=\left(x^{2} - 9\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

Therefore,

$$\frac{x^{2}}{2} + {\color{red}{\int{\frac{9 x}{x^{2} - 9} d x}}} = \frac{x^{2}}{2} + {\color{red}{\int{\frac{9}{2 u} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{9}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

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

$$\frac{x^{2}}{2} + \frac{9 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{x^{2}}{2} + \frac{9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recall that $$$u=x^{2} - 9$$$:

$$\frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{{\color{red}{\left(x^{2} - 9\right)}}}\right| \right)}}{2}$$

Therefore,

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

Add the constant of integration:

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

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