If ∫x2−4x4+9x2+16dx=Atan−1⁡(f(x))+B, then the value  of f(10) is

I=∫x2−4x4+9x2+16dxDivide both numerator and denominator by x2I=∫1−4x2x2+9+16x2dx=∫1−4x2x+4x2+1dx Put u=x+4x,du=1−4x2dxI=∫duu2+1=tan−1⁡(u)+B=1⋅tan−1⁡x+4x+B⇒f(x)=x+4x