$$$\frac{2}{7 - x^{2}}$$$'nin integrali

Bulun: $$$\int \frac{2}{7 - x^{2}}\, dx$$$.

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=2$$$ ve $$$f{\left(x \right)} = \frac{1}{7 - x^{2}}$$$ ile uygula:

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

Kısmi kesirlere ayrıştırma yapın (adımlar » görülebilir):

$$2 {\color{red}{\int{\frac{1}{7 - x^{2}} d x}}} = 2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}}$$

Her terimin integralini alın:

$$2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}} = 2 {\color{red}{\left(- \int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x} + \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}\right)}}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{\sqrt{7}}{14}$$$ ve $$$f{\left(x \right)} = \frac{1}{x - \sqrt{7}}$$$ ile uygula:

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x}}} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x - \sqrt{7}} d x}}{14}\right)}}$$

$$$u=x - \sqrt{7}$$$ olsun.

Böylece $$$du=\left(x - \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (adımlar » görülebilir) ve $$$dx = du$$$ elde ederiz.

Dolayısıyla,

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x - \sqrt{7}} d x}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

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

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

Hatırlayın ki $$$u=x - \sqrt{7}$$$:

$$- \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} = - \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x - \sqrt{7}\right)}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{\sqrt{7}}{14}$$$ ve $$$f{\left(x \right)} = \frac{1}{x + \sqrt{7}}$$$ ile uygula:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}}} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x + \sqrt{7}} d x}}{14}\right)}}$$

$$$u=x + \sqrt{7}$$$ olsun.

Böylece $$$du=\left(x + \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (adımlar » görülebilir) ve $$$dx = du$$$ elde ederiz.

Dolayısıyla,

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x + \sqrt{7}} d x}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

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

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

Hatırlayın ki $$$u=x + \sqrt{7}$$$:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x + \sqrt{7}\right)}}}\right| \right)}}{7}$$

Dolayısıyla,

$$\int{\frac{2}{7 - x^{2}} d x} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{x + \sqrt{7}}\right| \right)}}{7}$$

Sadeleştirin:

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}+C$$

$$$\int \frac{2}{7 - x^{2}}\, dx = \frac{\sqrt{7} \left(- \ln\left(\left|{x - \sqrt{7}}\right|\right) + \ln\left(\left|{x + \sqrt{7}}\right|\right)\right)}{7} + C$$$A