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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