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

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

Simplify the integrand:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-3$$$ and $$$f{\left(x \right)} = \frac{x}{2 x - 1}$$$:

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

Rewrite the numerator of the integrand as $$$x=\frac{1}{2}\left(2 x - 1\right)+\frac{1}{2}$$$ and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\frac{1}{2}$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{2 x - 1}$$$:

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

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

Then $$$du=\left(2 x - 1\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recall that $$$u=2 x - 1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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