$\int \frac{1}{x\left(x^7+1\right)} d x$ is equal to

$\frac{1}{7} \log \left(\frac{x^7}{x^7+1}\right)+C$

We have,

$I =\int \frac{1}{x\left(x^7+1\right)} d x=\int \frac{x^6}{x^7\left(x^7+1\right)} d x=\frac{1}{7} \int \frac{1}{x^7\left(x^7+1\right)} d\left(x^7\right)$

$\Rightarrow I =\frac{1}{7} \int \frac{1}{t(t+1)} d t=\frac{1}{7} \int \frac{1}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} d t$

$\Rightarrow I=\frac{1}{7} \times \frac{1}{2\left(\frac{1}{2}\right)} \log \left(\frac{t+\frac{1}{2}-\frac{1}{2}}{t+\frac{1}{2}+\frac{1}{2}}\right)+C$

$\Rightarrow I=\frac{1}{7} \log \left(\frac{t}{t+1}\right)+C=\frac{1}{7} \log \left(\frac{x^7}{x^7+1}\right)+C$