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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 1x 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
x