Integral de $$$\frac{3 x^{4}}{- 2 x^{4} + x^{3}}$$$

Encontre $$$\int \frac{3 x^{4}}{- 2 x^{4} + x^{3}}\, dx$$$.

Simplifique o integrando:

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

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

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

Reescreva o numerador do integrando como $$$x=\frac{1}{2}\left(2 x - 1\right)+\frac{1}{2}$$$ e decomponha a fração:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=\frac{1}{2}$$$:

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

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

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

Seja $$$u=2 x - 1$$$.

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

Portanto,

$$- \frac{3 x}{2} - \frac{3 {\color{red}{\int{\frac{1}{2 x - 1} d x}}}}{2} = - \frac{3 x}{2} - \frac{3 {\color{red}{\int{\frac{1}{2 u} d u}}}}{2}$$

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 $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

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

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

Recorde que $$$u=2 x - 1$$$:

$$- \frac{3 x}{2} - \frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = - \frac{3 x}{2} - \frac{3 \ln{\left(\left|{{\color{red}{\left(2 x - 1\right)}}}\right| \right)}}{4}$$

Portanto,

$$\int{\frac{3 x^{4}}{- 2 x^{4} + x^{3}} d x} = - \frac{3 x}{2} - \frac{3 \ln{\left(\left|{2 x - 1}\right| \right)}}{4}$$

Simplifique:

$$\int{\frac{3 x^{4}}{- 2 x^{4} + x^{3}} d x} = - \frac{3 \left(2 x + \ln{\left(\left|{2 x - 1}\right| \right)}\right)}{4}$$

Adicione a constante de integração:

$$\int{\frac{3 x^{4}}{- 2 x^{4} + x^{3}} d x} = - \frac{3 \left(2 x + \ln{\left(\left|{2 x - 1}\right| \right)}\right)}{4}+C$$

$$$\int \frac{3 x^{4}}{- 2 x^{4} + x^{3}}\, dx = - \frac{3 \left(2 x + \ln\left(\left|{2 x - 1}\right|\right)\right)}{4} + C$$$A