$$$11 x + \frac{17}{2 x^{2} + 7 x - 4}$$$ 的積分

求$$$\int \left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x}}} = {\color{red}{\left(\int{11 x d x} + \int{\frac{17}{2 x^{2} + 7 x - 4} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=11$$$ 與 $$$f{\left(x \right)} = x$$$：

$$\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + {\color{red}{\int{11 x d x}}} = \int{\frac{17}{2 x^{2} + 7 x - 4} d x} + {\color{red}{\left(11 \int{x d x}\right)}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

$$\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\int{x d x}}}=\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=17$$$ 與 $$$f{\left(x \right)} = \frac{1}{2 x^{2} + 7 x - 4}$$$：

$$\frac{11 x^{2}}{2} + {\color{red}{\int{\frac{17}{2 x^{2} + 7 x - 4} d x}}} = \frac{11 x^{2}}{2} + {\color{red}{\left(17 \int{\frac{1}{2 x^{2} + 7 x - 4} d x}\right)}}$$

進行部分分式分解（步驟可見 »）:

$$\frac{11 x^{2}}{2} + 17 {\color{red}{\int{\frac{1}{2 x^{2} + 7 x - 4} d x}}} = \frac{11 x^{2}}{2} + 17 {\color{red}{\int{\left(\frac{2}{9 \left(2 x - 1\right)} - \frac{1}{9 \left(x + 4\right)}\right)d x}}}$$

逐項積分：

$$\frac{11 x^{2}}{2} + 17 {\color{red}{\int{\left(\frac{2}{9 \left(2 x - 1\right)} - \frac{1}{9 \left(x + 4\right)}\right)d x}}} = \frac{11 x^{2}}{2} + 17 {\color{red}{\left(- \int{\frac{1}{9 \left(x + 4\right)} d x} + \int{\frac{2}{9 \left(2 x - 1\right)} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{9}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x + 4}$$$：

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - 17 {\color{red}{\int{\frac{1}{9 \left(x + 4\right)} d x}}} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - 17 {\color{red}{\left(\frac{\int{\frac{1}{x + 4} d x}}{9}\right)}}$$

令 $$$u=x + 4$$$。

則 $$$du=\left(x + 4\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{x + 4} d x}}}}{9} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{9}$$

回顧一下 $$$u=x + 4$$$：

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{9} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{{\color{red}{\left(x + 4\right)}}}\right| \right)}}{9} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{2}{9}$$$ 與 $$$f{\left(x \right)} = \frac{1}{2 x - 1}$$$：

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + 17 {\color{red}{\int{\frac{2}{9 \left(2 x - 1\right)} d x}}} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + 17 {\color{red}{\left(\frac{2 \int{\frac{1}{2 x - 1} d x}}{9}\right)}}$$

令 $$$u=2 x - 1$$$。

則 $$$du=\left(2 x - 1\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

該積分變為

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 x - 1} d x}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 u} d u}}}}{9}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 u} d u}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{9}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{9}$$

回顧一下 $$$u=2 x - 1$$$：

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{{\color{red}{\left(2 x - 1\right)}}}\right| \right)}}{9}$$

因此，

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{2 x - 1}\right| \right)}}{9}$$

化簡：

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{99 x^{2} - 34 \ln{\left(\left|{x + 4}\right| \right)} + 34 \ln{\left(\left|{2 x - 1}\right| \right)}}{18}$$

加上積分常數：

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{99 x^{2} - 34 \ln{\left(\left|{x + 4}\right| \right)} + 34 \ln{\left(\left|{2 x - 1}\right| \right)}}{18}+C$$

$$$\int \left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)\, dx = \frac{99 x^{2} - 34 \ln\left(\left|{x + 4}\right|\right) + 34 \ln\left(\left|{2 x - 1}\right|\right)}{18} + C$$$A