$$$\frac{9}{x^{3} \left(3 x - 2\right)}$$$'nin integrali

Bulun: $$$\int \frac{9}{x^{3} \left(3 x - 2\right)}\, dx$$$.

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=9$$$ ve $$$f{\left(x \right)} = \frac{1}{x^{3} \left(3 x - 2\right)}$$$ ile uygula:

$${\color{red}{\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x}}} = {\color{red}{\left(9 \int{\frac{1}{x^{3} \left(3 x - 2\right)} d x}\right)}}$$

Kısmi kesirlere ayrıştırma yapın (adımlar » görülebilir):

$$9 {\color{red}{\int{\frac{1}{x^{3} \left(3 x - 2\right)} d x}}} = 9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}}$$

Her terimin integralini alın:

$$9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}} = 9 {\color{red}{\left(- \int{\frac{1}{2 x^{3}} d x} - \int{\frac{3}{4 x^{2}} d x} - \int{\frac{9}{8 x} d x} + \int{\frac{27}{8 \left(3 x - 2\right)} d x}\right)}}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{9}{8}$$$ ve $$$f{\left(x \right)} = \frac{1}{x}$$$ ile uygula:

$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{9}{8 x} d x}}} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{9 \int{\frac{1}{x} d x}}{8}\right)}}$$

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

$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\int{\frac{1}{x} d x}}}}{8} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{8}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{3}{4}$$$ ve $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ ile uygula:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{3}{4 x^{2}} d x}}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{3 \int{\frac{1}{x^{2}} d x}}{4}\right)}}$$

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=-2$$$ ile uygulayın:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{\frac{1}{x^{2}} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{x^{-2} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- x^{-1}\right)}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- \frac{1}{x}\right)}}}{4}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$ ile uygula:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{1}{2 x^{3}} d x}}} + \frac{27}{4 x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{\int{\frac{1}{x^{3}} d x}}{2}\right)}} + \frac{27}{4 x}$$

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=-3$$$ ile uygulayın:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{\frac{1}{x^{3}} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{x^{-3} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{2} + \frac{27}{4 x}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{27}{8}$$$ ve $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$ ile uygula:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\int{\frac{27}{8 \left(3 x - 2\right)} d x}}} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\left(\frac{27 \int{\frac{1}{3 x - 2} d x}}{8}\right)}} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

$$$u=3 x - 2$$$ olsun.

Böylece $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (adımlar » görülebilir) ve $$$dx = \frac{du}{3}$$$ elde ederiz.

O halde,

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{3}$$$ ve $$$f{\left(u \right)} = \frac{1}{u}$$$ ile uygula:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

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

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\int{\frac{1}{u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

Hatırlayın ki $$$u=3 x - 2$$$:

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

Dolayısıyla,

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{3 x - 2}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

Sadeleştirin:

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}+C$$

$$$\int \frac{9}{x^{3} \left(3 x - 2\right)}\, dx = \frac{9 \left(9 x^{2} \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{3 x - 2}\right|\right)\right) + 6 x + 2\right)}{8 x^{2}} + C$$$A