Integral of $$$\frac{1}{2 x \left(x^{4} + 1\right)}$$$

Find $$$\int \frac{1}{2 x \left(x^{4} + 1\right)}\, dx$$$.

Let $$$u=x^{4} + 1$$$.

Then $$$du=\left(x^{4} + 1\right)^{\prime }dx = 4 x^{3} dx$$$ (steps can be seen »), and we have that $$$x^{3} dx = \frac{du}{4}$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{1}{2 x \left(x^{4} + 1\right)} d x}}} = {\color{red}{\int{\frac{1}{8 u \left(u - 1\right)} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(u \right)} = \frac{1}{u \left(u - 1\right)}$$$:

$${\color{red}{\int{\frac{1}{8 u \left(u - 1\right)} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u \left(u - 1\right)} d u}}{8}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u \left(u - 1\right)} d u}}}}{8} = \frac{{\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}}{8}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}}{8} = \frac{{\color{red}{\left(- \int{\frac{1}{u} d u} + \int{\frac{1}{u - 1} d u}\right)}}}{8}$$

Let $$$v=u - 1$$$.

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

Thus,

$$- \frac{\int{\frac{1}{u} d u}}{8} + \frac{{\color{red}{\int{\frac{1}{u - 1} d u}}}}{8} = - \frac{\int{\frac{1}{u} d u}}{8} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{8}$$

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

$$- \frac{\int{\frac{1}{u} d u}}{8} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{8} = - \frac{\int{\frac{1}{u} d u}}{8} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{8}$$

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

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{8} - \frac{\int{\frac{1}{u} d u}}{8} = \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{8} - \frac{\int{\frac{1}{u} d u}}{8}$$

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

$$\frac{\ln{\left(\left|{u - 1}\right| \right)}}{8} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\ln{\left(\left|{u - 1}\right| \right)}}{8} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recall that $$$u=x^{4} + 1$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{8} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = \frac{\ln{\left(\left|{-1 + {\color{red}{\left(x^{4} + 1\right)}}}\right| \right)}}{8} - \frac{\ln{\left(\left|{{\color{red}{\left(x^{4} + 1\right)}}}\right| \right)}}{8}$$

Therefore,

$$\int{\frac{1}{2 x \left(x^{4} + 1\right)} d x} = \frac{\ln{\left(x^{4} \right)}}{8} - \frac{\ln{\left(x^{4} + 1 \right)}}{8}$$

Simplify:

$$\int{\frac{1}{2 x \left(x^{4} + 1\right)} d x} = \frac{\ln{\left(x \right)}}{2} - \frac{\ln{\left(x^{4} + 1 \right)}}{8}$$

Add the constant of integration:

$$\int{\frac{1}{2 x \left(x^{4} + 1\right)} d x} = \frac{\ln{\left(x \right)}}{2} - \frac{\ln{\left(x^{4} + 1 \right)}}{8}+C$$

$$$\int \frac{1}{2 x \left(x^{4} + 1\right)}\, dx = \left(\frac{\ln\left(x\right)}{2} - \frac{\ln\left(x^{4} + 1\right)}{8}\right) + C$$$A