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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