Let

⇒ 2 = A(1 – x²) + (Bx + C)(1 – x)

⇒ 2 = A + Ax2 + Bx – Bx2 + C – Cx

On comparing the coefficients of x², x and constant term, we get,

A – B = 0

B – C = 0

A + C = 0

On solving these equations, we get,

A = 1, B =1 and C = 1

Thus,