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,







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