∫x4x2+1dx is equal to
x33+x+tan−1x+C
x33−x+tan−1+C
x22−x+2tan−1x+C
x33−x−tan−1x+C
Since degree of numerator is greater than the
degree of denominator, we divide them
x4x2+1=x2−1+1x2+1
so ∫x4x2+1dx=x33−x+tan−1x+C