2 x d x ^4 x+ ^4 x = ⁻¹(f(x))+c , then f ( 3 )=

Step-by-step solution

Let aligned I & = 2 x d x ^4 x+ ^4 x & = 2 x x ^4 x+ ^4 x d x aligned Dividing by ^2 x in numerator and denominator, we get = 2 x ^2 x 1+ ( ^2 x )^2 d x Again let ^2 x=t aligned I & = d t 1+t^2 & = ⁻¹ t+C & = ⁻¹ ( ^2 x )+C aligned By comparing with ⁻¹ f(x)+C , we get aligned f(x) & = ^2 x f ( 3 ) & = ( 3 ) ^2=( 3 )^2=3 . aligned