Evaluate the following integral:



x2 = (Ax + B)(3x2 + 4) + (Cx + D)(x2 + 1)


= (3A + C) x3 + (3B + D)x2 + (4A + C)x + 4B + D


Equating similar terms


3A + C = 0


3B + D = 1


4A + C = 0


4B + D = 0


Solving we get,


A = 0, B = – 1, C = 0,D = 4


Thus,






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