∫ x 2 + 1 e x x + 1 2 d x = f x e x + C , where C is a constant, then d 3 f d x 3 at x = 1 is equal to

Step-by-step solution

∫ x 2 + 1 e x d x x + 1 2 = f x e x + C ∫ e x x 2 - 1 x + 1 2 + 2 x + 1 2 d x = f x e x + C ∫ e x x - 1 x + 1 + 2 x + 1 2 d x = f x e x + C We know that ∫ e x f x + f ' x = e x f x + c Here f x = x - 1 x + 1   &   f ' x = 2 x + 1 2 So ∫ e x x - 1 x + 1 + 2 ( x + 1 ) 2 d x = f x e x + C ⇒ e x x - 1 x + 1 + C = e x f x + C On comparing both sides we get f x = x - 1 x + 1 So f ' x = 2 x + 1 2   &   f " x = - 4 x + 1 3 f ''' x = 12