How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$? I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?

$$\frac{1}{x^8(1+x^2)}=\frac{1+x^2-x^2}{x^8(1+x^2)}=\frac{1}{x^8}-\frac{1}{x^6(1+x^2)}=\cdots$$
and so on, finally you will get:
$$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$
