If d x ^4 x+ ^4 x = 1 2 ⁻¹[g(x)]+C , then g(x) equals

Step-by-step solution

aligned & (a) I= d x ^4 x+ ^4 x & = ^4 x 1+ ^4 x d x= ^2 x (1+ ^2 x ) 1+ ^4 x d x aligned Let x=t , then ^2 x d x=d t aligned I & = 1+t^2 1+t^4 d t= (1+ 1 t^2 ) (t^2+ 1 t^2 ) d t & = (1+ (1 / t^2 ) ) (t- 1 t )^2+2 d t aligned Let t- 1 t =u , then (1+ 1 t^2 ) d t=d u aligned I & = d u u^2+2 = d u u^2+( 2 )^2 & = 1 2 ⁻¹ ( u 2 )+C I & = 1 2 ⁻¹ ( t- 1 t 2 )+C & = 1 2 ⁻¹ ( t^2-1 2 t )+C I & = 1 2 ⁻¹ ( ^2 x-1 2 x )+C g(x) & = ^2 x-1 2 x = 1 2 ( x- x) aligned