2 x 1-x^2-x^4 d x= ⁻¹[f(x)]+c then f(x)=

Step-by-step solution

Let I = 2x 1 - x^2 - x^4 dx . Substitute t = x^2 , so dt = 2x dx . The integral becomes I = dt 1 - t - t^2 . Complete the square in the denominator: 1 - t - t^2 = 1 - (t^2 + t) = 1 - (t^2 + t + 1 4 - 1 4 ) = 1 + 1 4 - (t + 1 2 )^2 = 5 4 - (t + 1 2 )^2 . Thus, I = dt ( 5 2 )^2 - (t + 1 2 )^2 . Using the standard integral du a^2 - u^2 = ⁻¹( u a ) + C , we get I = ⁻¹( t + 1/2 5 /2 ) + C . Simplifying the argument: t + 1/2 5 /2 = 2t + 1 5 . Substituting t = x^2 back, I = ⁻¹( 2x^2 + 1 5 ) + C . Comparing with ⁻¹[f(x)] +