By dividing the two terms in the numerator by
x, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand:
xx2+1dx=xx2+x1dx=x+x−1dx=2x2+lnx+C
where
C
is an arbitrary constant.
NOTE: observe that
x1
has a singularity at
x=0, so the answers above are only primitive functions over intervals that do not contain
x=0.
xx2+1dx=
xx2+x1
dx=
x+x−1
dx=2x2+ln
x