Integral of $$$x^{6} + 3 x^{2} - 1$$$
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Find $$$\int \left(x^{6} + 3 x^{2} - 1\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(x^{6} + 3 x^{2} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{3 x^{2} d x} + \int{x^{6} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\int{3 x^{2} d x} + \int{x^{6} d x} - {\color{red}{\int{1 d x}}} = \int{3 x^{2} d x} + \int{x^{6} d x} - {\color{red}{x}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=6$$$:
$$- x + \int{3 x^{2} d x} + {\color{red}{\int{x^{6} d x}}}=- x + \int{3 x^{2} d x} + {\color{red}{\frac{x^{1 + 6}}{1 + 6}}}=- x + \int{3 x^{2} d x} + {\color{red}{\left(\frac{x^{7}}{7}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = x^{2}$$$:
$$\frac{x^{7}}{7} - x + {\color{red}{\int{3 x^{2} d x}}} = \frac{x^{7}}{7} - x + {\color{red}{\left(3 \int{x^{2} d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$\frac{x^{7}}{7} - x + 3 {\color{red}{\int{x^{2} d x}}}=\frac{x^{7}}{7} - x + 3 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{7}}{7} - x + 3 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
Therefore,
$$\int{\left(x^{6} + 3 x^{2} - 1\right)d x} = \frac{x^{7}}{7} + x^{3} - x$$
Add the constant of integration:
$$\int{\left(x^{6} + 3 x^{2} - 1\right)d x} = \frac{x^{7}}{7} + x^{3} - x+C$$
Answer
$$$\int \left(x^{6} + 3 x^{2} - 1\right)\, dx = \left(\frac{x^{7}}{7} + x^{3} - x\right) + C$$$A