Integral de $$$\frac{1}{6 x^{3} - 7 x^{2} - 3 x}$$$

Encontre $$$\int \frac{1}{6 x^{3} - 7 x^{2} - 3 x}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{6 x^{3} - 7 x^{2} - 3 x} d x}}} = {\color{red}{\int{\left(\frac{9}{11 \left(3 x + 1\right)} + \frac{4}{33 \left(2 x - 3\right)} - \frac{1}{3 x}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(\frac{9}{11 \left(3 x + 1\right)} + \frac{4}{33 \left(2 x - 3\right)} - \frac{1}{3 x}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{3 x} d x} + \int{\frac{4}{33 \left(2 x - 3\right)} d x} + \int{\frac{9}{11 \left(3 x + 1\right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{3}$$$ e $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$\int{\frac{4}{33 \left(2 x - 3\right)} d x} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} - {\color{red}{\int{\frac{1}{3 x} d x}}} = \int{\frac{4}{33 \left(2 x - 3\right)} d x} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{3}\right)}}$$

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

$$\int{\frac{4}{33 \left(2 x - 3\right)} d x} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{3} = \int{\frac{4}{33 \left(2 x - 3\right)} d x} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{3}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{4}{33}$$$ e $$$f{\left(x \right)} = \frac{1}{2 x - 3}$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + {\color{red}{\int{\frac{4}{33 \left(2 x - 3\right)} d x}}} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + {\color{red}{\left(\frac{4 \int{\frac{1}{2 x - 3} d x}}{33}\right)}}$$

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

Então $$$du=\left(2 x - 3\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.

A integral torna-se

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{4 {\color{red}{\int{\frac{1}{2 x - 3} d x}}}}{33} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{4 {\color{red}{\int{\frac{1}{2 u} d u}}}}{33}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{4 {\color{red}{\int{\frac{1}{2 u} d u}}}}{33} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{4 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{33}$$

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

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{2 {\color{red}{\int{\frac{1}{u} d u}}}}{33} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} + \frac{2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{33}$$

Recorde que $$$u=2 x - 3$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{33} + \int{\frac{9}{11 \left(3 x + 1\right)} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{{\color{red}{\left(2 x - 3\right)}}}\right| \right)}}{33} + \int{\frac{9}{11 \left(3 x + 1\right)} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{9}{11}$$$ e $$$f{\left(x \right)} = \frac{1}{3 x + 1}$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + {\color{red}{\int{\frac{9}{11 \left(3 x + 1\right)} d x}}} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + {\color{red}{\left(\frac{9 \int{\frac{1}{3 x + 1} d x}}{11}\right)}}$$

Seja $$$u=3 x + 1$$$.

Então $$$du=\left(3 x + 1\right)^{\prime }dx = 3 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{3}$$$.

Logo,

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{9 {\color{red}{\int{\frac{1}{3 x + 1} d x}}}}{11} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{9 {\color{red}{\int{\frac{1}{3 u} d u}}}}{11}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{3}$$$ e $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{9 {\color{red}{\int{\frac{1}{3 u} d u}}}}{11} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{9 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{11}$$

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

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{11} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{11}$$

Recorde que $$$u=3 x + 1$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{11} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{3 \ln{\left(\left|{{\color{red}{\left(3 x + 1\right)}}}\right| \right)}}{11}$$

Portanto,

$$\int{\frac{1}{6 x^{3} - 7 x^{2} - 3 x} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{2 \ln{\left(\left|{2 x - 3}\right| \right)}}{33} + \frac{3 \ln{\left(\left|{3 x + 1}\right| \right)}}{11}$$

Simplifique:

$$\int{\frac{1}{6 x^{3} - 7 x^{2} - 3 x} d x} = \frac{- 11 \ln{\left(\left|{x}\right| \right)} + 2 \ln{\left(\left|{2 x - 3}\right| \right)} + 9 \ln{\left(\left|{3 x + 1}\right| \right)}}{33}$$

Adicione a constante de integração:

$$\int{\frac{1}{6 x^{3} - 7 x^{2} - 3 x} d x} = \frac{- 11 \ln{\left(\left|{x}\right| \right)} + 2 \ln{\left(\left|{2 x - 3}\right| \right)} + 9 \ln{\left(\left|{3 x + 1}\right| \right)}}{33}+C$$

$$$\int \frac{1}{6 x^{3} - 7 x^{2} - 3 x}\, dx = \frac{- 11 \ln\left(\left|{x}\right|\right) + 2 \ln\left(\left|{2 x - 3}\right|\right) + 9 \ln\left(\left|{3 x + 1}\right|\right)}{33} + C$$$A