$$$\frac{1}{1 - x^{2}}$$$'nin integrali

Bulun: $$$\int \frac{1}{1 - x^{2}}\, dx$$$.

Kısmi kesirlere ayrıştırma yapın (adımlar » görülebilir):

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

Her terimin integralini alın:

$${\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{2 \left(x - 1\right)} d x} + \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \frac{1}{x + 1}$$$ ile uygula:

$$- \int{\frac{1}{2 \left(x - 1\right)} d x} + {\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}} = - \int{\frac{1}{2 \left(x - 1\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}$$

$$$u=x + 1$$$ olsun.

Böylece $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (adımlar » görülebilir) ve $$$dx = du$$$ elde ederiz.

Dolayısıyla,

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

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

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

Hatırlayın ki $$$u=x + 1$$$:

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

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \frac{1}{x - 1}$$$ ile uygula:

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

$$$u=x - 1$$$ olsun.

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

Dolayısıyla,

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

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

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

Hatırlayın ki $$$u=x - 1$$$:

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

Dolayısıyla,

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

Sadeleştirin:

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

İntegrasyon sabitini ekleyin:

$$\int{\frac{1}{1 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{2}+C$$

$$$\int \frac{1}{1 - x^{2}}\, dx = \frac{- \ln\left(\left|{x - 1}\right|\right) + \ln\left(\left|{x + 1}\right|\right)}{2} + C$$$A