Using partial fraction decomposition, we can write the integrand as

- ${\frac {1}{x^{4}+x^{2}}}={\frac {1}{x^{2}(x^{2}+1)}}={\frac {A}{x}}+{\frac {B}{x^{2}}}+{\frac {Cx+D}{x^{2}+1}}$

for some constants $A,B,C,D$. We find that $A=0,B=1,C=0,D=-1$, which implies that

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

The first integral on the right-hand side of the equation can be easily evaluated:

- $\int {\frac {1}{x^{2}}}\,dx=-{\frac {1}{x}}+C.$

On the other hand, to compute the second integral, we can use the substitution $x=\tan u$.
Then $1+x^{2}=1+\tan ^{2}u=\sec ^{2}u$ and $dx=\sec ^{2}u\,du$, which gives

- $\int {\frac {1}{1+x^{2}}}\,dx=\int {\frac {1}{\sec ^{2}u}}\sec ^{2}u\,du=\int 1\,du=u+C=\arctan x+C.$

Combining these, we obtain

- $\int {\frac {1}{x^{4}+x^{2}}}dx=\color {blue}-{\frac {1}{x}}-\arctan x+C.$