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