$$$n$$$ değişkenine göre $$$\frac{1}{p \left(1 - \frac{p}{n}\right)}$$$ fonksiyonunun integrali

Bulun: $$$\int \frac{1}{p \left(1 - \frac{p}{n}\right)}\, dn$$$.

Sabit katsayı kuralı $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$'i $$$c=\frac{1}{p}$$$ ve $$$f{\left(n \right)} = \frac{1}{1 - \frac{p}{n}}$$$ ile uygula:

$${\color{red}{\int{\frac{1}{p \left(1 - \frac{p}{n}\right)} d n}}} = {\color{red}{\frac{\int{\frac{1}{1 - \frac{p}{n}} d n}}{p}}}$$

Simplify:

$$\frac{{\color{red}{\int{\frac{1}{1 - \frac{p}{n}} d n}}}}{p} = \frac{{\color{red}{\int{\frac{n}{n - p} d n}}}}{p}$$

Kesri yeniden yazın ve parçalara ayırın:

$$\frac{{\color{red}{\int{\frac{n}{n - p} d n}}}}{p} = \frac{{\color{red}{\int{\left(\frac{p}{n - p} + 1\right)d n}}}}{p}$$

Her terimin integralini alın:

$$\frac{{\color{red}{\int{\left(\frac{p}{n - p} + 1\right)d n}}}}{p} = \frac{{\color{red}{\left(\int{1 d n} + \int{\frac{p}{n - p} d n}\right)}}}{p}$$

$$$c=1$$$ kullanarak $$$\int c\, dn = c n$$$ sabit kuralını uygula:

$$\frac{\int{\frac{p}{n - p} d n} + {\color{red}{\int{1 d n}}}}{p} = \frac{\int{\frac{p}{n - p} d n} + {\color{red}{n}}}{p}$$

Sabit katsayı kuralı $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$'i $$$c=p$$$ ve $$$f{\left(n \right)} = \frac{1}{n - p}$$$ ile uygula:

$$\frac{n + {\color{red}{\int{\frac{p}{n - p} d n}}}}{p} = \frac{n + {\color{red}{p \int{\frac{1}{n - p} d n}}}}{p}$$

$$$u=n - p$$$ olsun.

Böylece $$$du=\left(n - p\right)^{\prime }dn = 1 dn$$$ (adımlar » görülebilir) ve $$$dn = du$$$ elde ederiz.

İntegral şu şekilde yeniden yazılabilir:

$$\frac{n + p {\color{red}{\int{\frac{1}{n - p} d n}}}}{p} = \frac{n + p {\color{red}{\int{\frac{1}{u} d u}}}}{p}$$

$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{n + p {\color{red}{\int{\frac{1}{u} d u}}}}{p} = \frac{n + p {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{p}$$

Hatırlayın ki $$$u=n - p$$$:

$$\frac{n + p \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{p} = \frac{n + p \ln{\left(\left|{{\color{red}{\left(n - p\right)}}}\right| \right)}}{p}$$

Dolayısıyla,

$$\int{\frac{1}{p \left(1 - \frac{p}{n}\right)} d n} = \frac{n + p \ln{\left(\left|{n - p}\right| \right)}}{p}$$

Sadeleştirin:

$$\int{\frac{1}{p \left(1 - \frac{p}{n}\right)} d n} = \frac{n}{p} + \ln{\left(\left|{n - p}\right| \right)}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{1}{p \left(1 - \frac{p}{n}\right)} d n} = \frac{n}{p} + \ln{\left(\left|{n - p}\right| \right)}+C$$

$$$\int \frac{1}{p \left(1 - \frac{p}{n}\right)}\, dn = \left(\frac{n}{p} + \ln\left(\left|{n - p}\right|\right)\right) + C$$$A