If f(x) is a polynomial satisfying f(x)⋅f1x=f(x)+f1x and f(3)=82, then ∫f(x)x2+1dx=
x3−x+2tan−1x+c
13x3−x+tan−1x+c
x33−x+2tan−1x+c
13x3+x+2tan−1x+c
Let f(x)=xn+1f(3)=82⇒3n+1=82⇒3n=34∴n=4∫x4+1x2+1dx=∫x2−1+2x2+1dx