Let f(x) be a polynomial of degree three such that f(0)=1,f(1)=2 and 0 is a critical point of f(x)such that f(x) does not have a local extremum at 0. Then∫f(x)x2+1dx is equal to

Let f(x)=ax3+bx2+cx+d. Sincef(0)=1 so d=1. Moreover,  f(1)=2a + b + c = 1 since 0 is a critical point, 0 =f′(0)=3a.0+2b.0+C .i.e.,  C=0.  Also  f′′(x)=6ax+2band sincef(x) does not have extremum at 0 so 0 =f′′(0)=b.Hence a = 1  and ∫f(x)x2+1dx=∫x3+1x2+1dx=∫x−122xx2+1+1x2+1dx=12x2−log⁡x2+1+tan−1⁡x+C.