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