Using partial fraction decomposition, we can write the integrand as
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d30325cd95245fd653c28e33e9e9d849cddc75fd)
for some constants
. We find that
, which implies that
![{\displaystyle \int {\frac {1}{x^{4}+x^{2}}}\,dx=\int {\frac {1}{x^{2}}}\,dx-\int {\frac {1}{1+x^{2}}}\,dx.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/92edca82a726401298520ec399640e9760abf5a4)
The first integral on the right-hand side of the equation can be easily evaluated:
![{\displaystyle \int {\frac {1}{x^{2}}}\,dx=-{\frac {1}{x}}+C.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a652afbb6489cbee5696e3b896e89ec07830ac02)
On the other hand, to compute the second integral, we can use the substitution
.
Then
and
, which gives
![{\displaystyle \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.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cf0bd6b5baac7fc17416b1177b3ea56e3681bd89)
Combining these, we obtain
![{\displaystyle \int {\frac {1}{x^{4}+x^{2}}}dx=\color {blue}-{\frac {1}{x}}-\arctan x+C.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7ac2ca666df57c2104b30bcc08228f7249aefb55)