Integral of $$$\frac{3 x - 1}{x^{4}}$$$
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Find $$$\int \frac{3 x - 1}{x^{4}}\, dx$$$.
Solution
Expand the expression:
$${\color{red}{\int{\frac{3 x - 1}{x^{4}} d x}}} = {\color{red}{\int{\left(\frac{3}{x^{3}} - \frac{1}{x^{4}}\right)d x}}}$$
Integrate term by term:
$${\color{red}{\int{\left(\frac{3}{x^{3}} - \frac{1}{x^{4}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x^{4}} d x} + \int{\frac{3}{x^{3}} d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:
$$\int{\frac{3}{x^{3}} d x} - {\color{red}{\int{\frac{1}{x^{4}} d x}}}=\int{\frac{3}{x^{3}} d x} - {\color{red}{\int{x^{-4} d x}}}=\int{\frac{3}{x^{3}} d x} - {\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}=\int{\frac{3}{x^{3}} d x} - {\color{red}{\left(- \frac{x^{-3}}{3}\right)}}=\int{\frac{3}{x^{3}} d x} - {\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:
$${\color{red}{\int{\frac{3}{x^{3}} d x}}} + \frac{1}{3 x^{3}} = {\color{red}{\left(3 \int{\frac{1}{x^{3}} d x}\right)}} + \frac{1}{3 x^{3}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:
$$3 {\color{red}{\int{\frac{1}{x^{3}} d x}}} + \frac{1}{3 x^{3}}=3 {\color{red}{\int{x^{-3} d x}}} + \frac{1}{3 x^{3}}=3 {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}} + \frac{1}{3 x^{3}}=3 {\color{red}{\left(- \frac{x^{-2}}{2}\right)}} + \frac{1}{3 x^{3}}=3 {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}} + \frac{1}{3 x^{3}}$$
Therefore,
$$\int{\frac{3 x - 1}{x^{4}} d x} = - \frac{3}{2 x^{2}} + \frac{1}{3 x^{3}}$$
Simplify:
$$\int{\frac{3 x - 1}{x^{4}} d x} = \frac{2 - 9 x}{6 x^{3}}$$
Add the constant of integration:
$$\int{\frac{3 x - 1}{x^{4}} d x} = \frac{2 - 9 x}{6 x^{3}}+C$$
Answer
$$$\int \frac{3 x - 1}{x^{4}}\, dx = \frac{2 - 9 x}{6 x^{3}} + C$$$A