$$$\left(x - 2\right)^{4} \left(x - 1\right)^{3}$$$ 的积分
您的输入
求$$$\int \left(x - 2\right)^{4} \left(x - 1\right)^{3}\, dx$$$。
解答
设$$$u=x - 2$$$。
则$$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。
积分变为
$${\color{red}{\int{\left(x - 2\right)^{4} \left(x - 1\right)^{3} d x}}} = {\color{red}{\int{u^{4} \left(u + 1\right)^{3} d u}}}$$
Expand the expression:
$${\color{red}{\int{u^{4} \left(u + 1\right)^{3} d u}}} = {\color{red}{\int{\left(u^{7} + 3 u^{6} + 3 u^{5} + u^{4}\right)d u}}}$$
逐项积分：
$${\color{red}{\int{\left(u^{7} + 3 u^{6} + 3 u^{5} + u^{4}\right)d u}}} = {\color{red}{\left(\int{u^{4} d u} + \int{3 u^{5} d u} + \int{3 u^{6} d u} + \int{u^{7} d u}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=4$$$：
$$\int{3 u^{5} d u} + \int{3 u^{6} d u} + \int{u^{7} d u} + {\color{red}{\int{u^{4} d u}}}=\int{3 u^{5} d u} + \int{3 u^{6} d u} + \int{u^{7} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\int{3 u^{5} d u} + \int{3 u^{6} d u} + \int{u^{7} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=7$$$：
$$\frac{u^{5}}{5} + \int{3 u^{5} d u} + \int{3 u^{6} d u} + {\color{red}{\int{u^{7} d u}}}=\frac{u^{5}}{5} + \int{3 u^{5} d u} + \int{3 u^{6} d u} + {\color{red}{\frac{u^{1 + 7}}{1 + 7}}}=\frac{u^{5}}{5} + \int{3 u^{5} d u} + \int{3 u^{6} d u} + {\color{red}{\left(\frac{u^{8}}{8}\right)}}$$
对 $$$c=3$$$ 和 $$$f{\left(u \right)} = u^{5}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$$\frac{u^{8}}{8} + \frac{u^{5}}{5} + \int{3 u^{6} d u} + {\color{red}{\int{3 u^{5} d u}}} = \frac{u^{8}}{8} + \frac{u^{5}}{5} + \int{3 u^{6} d u} + {\color{red}{\left(3 \int{u^{5} d u}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=5$$$：
$$\frac{u^{8}}{8} + \frac{u^{5}}{5} + \int{3 u^{6} d u} + 3 {\color{red}{\int{u^{5} d u}}}=\frac{u^{8}}{8} + \frac{u^{5}}{5} + \int{3 u^{6} d u} + 3 {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=\frac{u^{8}}{8} + \frac{u^{5}}{5} + \int{3 u^{6} d u} + 3 {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$
对 $$$c=3$$$ 和 $$$f{\left(u \right)} = u^{6}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$$\frac{u^{8}}{8} + \frac{u^{6}}{2} + \frac{u^{5}}{5} + {\color{red}{\int{3 u^{6} d u}}} = \frac{u^{8}}{8} + \frac{u^{6}}{2} + \frac{u^{5}}{5} + {\color{red}{\left(3 \int{u^{6} d u}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=6$$$：
$$\frac{u^{8}}{8} + \frac{u^{6}}{2} + \frac{u^{5}}{5} + 3 {\color{red}{\int{u^{6} d u}}}=\frac{u^{8}}{8} + \frac{u^{6}}{2} + \frac{u^{5}}{5} + 3 {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=\frac{u^{8}}{8} + \frac{u^{6}}{2} + \frac{u^{5}}{5} + 3 {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$
回忆一下 $$$u=x - 2$$$:
$$\frac{{\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{6}}{2} + \frac{3 {\color{red}{u}}^{7}}{7} + \frac{{\color{red}{u}}^{8}}{8} = \frac{{\color{red}{\left(x - 2\right)}}^{5}}{5} + \frac{{\color{red}{\left(x - 2\right)}}^{6}}{2} + \frac{3 {\color{red}{\left(x - 2\right)}}^{7}}{7} + \frac{{\color{red}{\left(x - 2\right)}}^{8}}{8}$$
因此，
$$\int{\left(x - 2\right)^{4} \left(x - 1\right)^{3} d x} = \frac{\left(x - 2\right)^{8}}{8} + \frac{3 \left(x - 2\right)^{7}}{7} + \frac{\left(x - 2\right)^{6}}{2} + \frac{\left(x - 2\right)^{5}}{5}$$
化简：
$$\int{\left(x - 2\right)^{4} \left(x - 1\right)^{3} d x} = \frac{\left(x - 2\right)^{5} \left(140 x + 35 \left(x - 2\right)^{3} + 120 \left(x - 2\right)^{2} - 224\right)}{280}$$
加上积分常数：
$$\int{\left(x - 2\right)^{4} \left(x - 1\right)^{3} d x} = \frac{\left(x - 2\right)^{5} \left(140 x + 35 \left(x - 2\right)^{3} + 120 \left(x - 2\right)^{2} - 224\right)}{280}+C$$
答案
$$$\int \left(x - 2\right)^{4} \left(x - 1\right)^{3}\, dx = \frac{\left(x - 2\right)^{5} \left(140 x + 35 \left(x - 2\right)^{3} + 120 \left(x - 2\right)^{2} - 224\right)}{280} + C$$$A