Integral de $$$\frac{x^{2} - 3}{x^{3} - 72 x}$$$

Encontre $$$\int \frac{x^{2} - 3}{x^{3} - 72 x}\, dx$$$.

Efetue a decomposição em frações parciais (os passos podem ser vistos »):

$${\color{red}{\int{\frac{x^{2} - 3}{x^{3} - 72 x} d x}}} = {\color{red}{\int{\left(\frac{23}{48 \left(x + 6 \sqrt{2}\right)} + \frac{23}{48 \left(x - 6 \sqrt{2}\right)} + \frac{1}{24 x}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(\frac{23}{48 \left(x + 6 \sqrt{2}\right)} + \frac{23}{48 \left(x - 6 \sqrt{2}\right)} + \frac{1}{24 x}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{24 x} d x} + \int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{24}$$$ e $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$\int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + {\color{red}{\int{\frac{1}{24 x} d x}}} = \int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{24}\right)}}$$

A integral de $$$\frac{1}{x}$$$ é $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

$$\int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{24} = \int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{24}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{23}{48}$$$ e $$$f{\left(x \right)} = \frac{1}{x - 6 \sqrt{2}}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + {\color{red}{\int{\frac{23}{48 \left(x - 6 \sqrt{2}\right)} d x}}} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + {\color{red}{\left(\frac{23 \int{\frac{1}{x - 6 \sqrt{2}} d x}}{48}\right)}}$$

Seja $$$u=x - 6 \sqrt{2}$$$.

Então $$$du=\left(x - 6 \sqrt{2}\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

A integral pode ser reescrita como

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{23 {\color{red}{\int{\frac{1}{x - 6 \sqrt{2}} d x}}}}{48} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{23 {\color{red}{\int{\frac{1}{u} d u}}}}{48}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{23 {\color{red}{\int{\frac{1}{u} d u}}}}{48} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} + \frac{23 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{48}$$

Recorde que $$$u=x - 6 \sqrt{2}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{48} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{{\color{red}{\left(x - 6 \sqrt{2}\right)}}}\right| \right)}}{48} + \int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{23}{48}$$$ e $$$f{\left(x \right)} = \frac{1}{x + 6 \sqrt{2}}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + {\color{red}{\int{\frac{23}{48 \left(x + 6 \sqrt{2}\right)} d x}}} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + {\color{red}{\left(\frac{23 \int{\frac{1}{x + 6 \sqrt{2}} d x}}{48}\right)}}$$

Seja $$$u=x + 6 \sqrt{2}$$$.

Então $$$du=\left(x + 6 \sqrt{2}\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Portanto,

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 {\color{red}{\int{\frac{1}{x + 6 \sqrt{2}} d x}}}}{48} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 {\color{red}{\int{\frac{1}{u} d u}}}}{48}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 {\color{red}{\int{\frac{1}{u} d u}}}}{48} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{48}$$

Recorde que $$$u=x + 6 \sqrt{2}$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{48} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 \ln{\left(\left|{{\color{red}{\left(x + 6 \sqrt{2}\right)}}}\right| \right)}}{48}$$

Portanto,

$$\int{\frac{x^{2} - 3}{x^{3} - 72 x} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{24} + \frac{23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)}}{48} + \frac{23 \ln{\left(\left|{x + 6 \sqrt{2}}\right| \right)}}{48}$$

Simplifique:

$$\int{\frac{x^{2} - 3}{x^{3} - 72 x} d x} = \frac{2 \ln{\left(\left|{x}\right| \right)} + 23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)} + 23 \ln{\left(\left|{x + 6 \sqrt{2}}\right| \right)}}{48}$$

Adicione a constante de integração:

$$\int{\frac{x^{2} - 3}{x^{3} - 72 x} d x} = \frac{2 \ln{\left(\left|{x}\right| \right)} + 23 \ln{\left(\left|{x - 6 \sqrt{2}}\right| \right)} + 23 \ln{\left(\left|{x + 6 \sqrt{2}}\right| \right)}}{48}+C$$

$$$\int \frac{x^{2} - 3}{x^{3} - 72 x}\, dx = \frac{2 \ln\left(\left|{x}\right|\right) + 23 \ln\left(\left|{x - 6 \sqrt{2}}\right|\right) + 23 \ln\left(\left|{x + 6 \sqrt{2}}\right|\right)}{48} + C$$$A