Integral of $$$\frac{2 x^{4}}{x^{4} - 1}$$$

Find $$$\int \frac{2 x^{4}}{x^{4} - 1}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \frac{x^{4}}{x^{4} - 1}$$$:

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$2 \int{\frac{1}{x^{4} - 1} d x} + 2 {\color{red}{\int{1 d x}}} = 2 \int{\frac{1}{x^{4} - 1} d x} + 2 {\color{red}{x}}$$

Perform partial fraction decomposition (steps can be seen »):

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

Integrate term by term:

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

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 $$$f{\left(x \right)} = \frac{1}{x^{2} + 1}$$$:

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

The integral of $$$\frac{1}{x^{2} + 1}$$$ is $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 1}$$$:

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

Let $$$u=x + 1$$$.

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

The integral becomes

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

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

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

Recall that $$$u=x + 1$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

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

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

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

The integral becomes

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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