The integral ∫3x13+2x11(2x4+3x2+1)4dx  is equal to (where C is a constant of integration)

LetI=∫  3x13+2x11(2x4+3x2+1)4dx=∫  3x3+2x5(2+3x2+1x4)−4dx[On dividing numerator and denominator by x16]Now, put 2+3x2+1x4=t⇒(−6x3−4x5)dx=dt⇒(3x3+2x5)dx=−dt2So, I=∫  −dt2t4=−12×t−4+1−4+1+C=16t3+C=16(2+3x2+1x4)3+C [∵ t=2+3x2+1x4] =x126(2x4+3x2+1)3+C