Integrate the rational functions.

Let

⇒3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C( x – 1)²

⇒3x + 5 = A(x² - 1) + B(x + 1) + C( x² + 1 – 2x) …(1)

Substituting x = 1 in equation (1), we get,

B = 4

Equating the coefficients of x² and x, we get,

A + C = 0

b – 2C = 3

A = and c =

Thus,