Integral of $$$9 x + \frac{6}{x}$$$
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Find $$$\int \left(9 x + \frac{6}{x}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(9 x + \frac{6}{x}\right)d x}}} = {\color{red}{\left(\int{\frac{6}{x} d x} + \int{9 x d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=6$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:
$$\int{9 x d x} + {\color{red}{\int{\frac{6}{x} d x}}} = \int{9 x d x} + {\color{red}{\left(6 \int{\frac{1}{x} d x}\right)}}$$
The integral of $$$\frac{1}{x}$$$ is $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:
$$\int{9 x d x} + 6 {\color{red}{\int{\frac{1}{x} d x}}} = \int{9 x d x} + 6 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=9$$$ and $$$f{\left(x \right)} = x$$$:
$$6 \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{9 x d x}}} = 6 \ln{\left(\left|{x}\right| \right)} + {\color{red}{\left(9 \int{x d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$6 \ln{\left(\left|{x}\right| \right)} + 9 {\color{red}{\int{x d x}}}=6 \ln{\left(\left|{x}\right| \right)} + 9 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=6 \ln{\left(\left|{x}\right| \right)} + 9 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Therefore,
$$\int{\left(9 x + \frac{6}{x}\right)d x} = \frac{9 x^{2}}{2} + 6 \ln{\left(\left|{x}\right| \right)}$$
Add the constant of integration:
$$\int{\left(9 x + \frac{6}{x}\right)d x} = \frac{9 x^{2}}{2} + 6 \ln{\left(\left|{x}\right| \right)}+C$$
Answer
$$$\int \left(9 x + \frac{6}{x}\right)\, dx = \left(\frac{9 x^{2}}{2} + 6 \ln\left(\left|{x}\right|\right)\right) + C$$$A