$$$6 x^{2} - 5 x - 4$$$ 的积分
您的输入
求$$$\int \left(6 x^{2} - 5 x - 4\right)\, dx$$$。
解答
逐项积分:
$${\color{red}{\int{\left(6 x^{2} - 5 x - 4\right)d x}}} = {\color{red}{\left(- \int{4 d x} - \int{5 x d x} + \int{6 x^{2} d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$,使用 $$$c=4$$$:
$$- \int{5 x d x} + \int{6 x^{2} d x} - {\color{red}{\int{4 d x}}} = - \int{5 x d x} + \int{6 x^{2} d x} - {\color{red}{\left(4 x\right)}}$$
对 $$$c=5$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- 4 x + \int{6 x^{2} d x} - {\color{red}{\int{5 x d x}}} = - 4 x + \int{6 x^{2} d x} - {\color{red}{\left(5 \int{x d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=1$$$:
$$- 4 x + \int{6 x^{2} d x} - 5 {\color{red}{\int{x d x}}}=- 4 x + \int{6 x^{2} d x} - 5 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 4 x + \int{6 x^{2} d x} - 5 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
对 $$$c=6$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- \frac{5 x^{2}}{2} - 4 x + {\color{red}{\int{6 x^{2} d x}}} = - \frac{5 x^{2}}{2} - 4 x + {\color{red}{\left(6 \int{x^{2} d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=2$$$:
$$- \frac{5 x^{2}}{2} - 4 x + 6 {\color{red}{\int{x^{2} d x}}}=- \frac{5 x^{2}}{2} - 4 x + 6 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \frac{5 x^{2}}{2} - 4 x + 6 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
因此,
$$\int{\left(6 x^{2} - 5 x - 4\right)d x} = 2 x^{3} - \frac{5 x^{2}}{2} - 4 x$$
化简:
$$\int{\left(6 x^{2} - 5 x - 4\right)d x} = \frac{x \left(4 x^{2} - 5 x - 8\right)}{2}$$
加上积分常数:
$$\int{\left(6 x^{2} - 5 x - 4\right)d x} = \frac{x \left(4 x^{2} - 5 x - 8\right)}{2}+C$$
答案
$$$\int \left(6 x^{2} - 5 x - 4\right)\, dx = \frac{x \left(4 x^{2} - 5 x - 8\right)}{2} + C$$$A