$$$\frac{1}{a - p}$$$ 對 $$$a$$$ 的積分
您的輸入
求$$$\int \frac{1}{a - p}\, da$$$。
解答
令 $$$u=a - p$$$。
則 $$$du=\left(a - p\right)^{\prime }da = 1 da$$$ (步驟見»),並可得 $$$da = du$$$。
該積分可改寫為
$${\color{red}{\int{\frac{1}{a - p} d a}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回顧一下 $$$u=a - p$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(a - p\right)}}}\right| \right)}$$
因此,
$$\int{\frac{1}{a - p} d a} = \ln{\left(\left|{a - p}\right| \right)}$$
加上積分常數:
$$\int{\frac{1}{a - p} d a} = \ln{\left(\left|{a - p}\right| \right)}+C$$
答案
$$$\int \frac{1}{a - p}\, da = \ln\left(\left|{a - p}\right|\right) + C$$$A