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Integration Using Partial Fraction is considered one the most difficult concept.
22 Questions around this concept.
If and then the ordered pair (A, B) is equal to : (where C is a constant of integration)
Evaluate
Let .If , then is equal to
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A rational function is of the form P(x) / Q(x), where P(x) and Q(x) are polynomial function and Q(x) ≠ 0.
Partial fraction method is used to integrate rational functions.
Partial fraction decomposition can be applied to a rational function only if deg (P(x)) < deg (Q(x)).
In the case when deg (P(x)) ≥ deg(Q(x)), we must first perform long division to rewrite the quotient in the form of . where deg (R(x)) < deg (Q(x)). We then do a partial fraction decomposition on .
Illustration on how to approach to integrals of rational functions of the form where, deg (P(x)) ≥ deg(Q(x))
If the rational function is proper (i.e. deg (P(x)) < deg (Q(x)) ), then three cases arises
Case 1:
When denominator (Q(x)) is expressed as the product of the non-repeated linear factor
Short cut method
A, B, C can now be determined by
Option 1 : comparing coefficients of different powers of x and the constant terms on both sides.
Option 2: Since the relation that we just obtained should held true for all x, we substitute those values of x that would straight way give us the required values of A, B and C. These values are obviously the roots of Q(x).
Case 2:
When denominator (Q(x)) is expressed as the product of the linear factor such that some of them are repeating (Linear and repeated)
Case 3:
When some of the factors in denominator (Q(x)) are quadratic but non-repeating
Corresponding to each quadratic factor ax2 + bx + c, assume the partial fraction of the type
, where A and B are constants to be determined by comparing coefficients of similar power of x in the numerator of both sides. or by substituting suitable values of x on both sides
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