(i) iIf a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

(iv)

Given:-

This is a linear differential equation, comparing it with

I.F = e^∫Pdx

= (x² + 1)²

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y(x² + 1)² = –∫ dx + c

⇒ y(x² + 1)² = –x + c