Evaluate the following integral:



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


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


Equating similar terms


A + C = 0


B + D = 0


4A + C = 0


4B + D = 1


We get,


Thus,




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