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| Integral
calculus |
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The
indefinite integral |
Integrating
rational functions
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Use
of the
partial fraction decomposition to integrate a proper rational
function
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Use
of the
partial fraction decomposition to integrate a proper rational
function examples
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Integrating
rational functions
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Recall
that a rational function is a ratio of two polynomials
written Pn(x)
/ Qm(x).
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To
integrate a rational function
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in
case n >
m, that is, if the
degree of P
is greater than or equal to degree of Q,
we first divide Pn(x)
by Qm(x)
to obtain the sum of the polynomial Pn
-
m(x)
and a proper or simple rational function.
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Then,
we integrate proper rational function using decomposition of rational function
into a sum of partial fractions.
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Use
of the
partial fraction decomposition to integrate a proper rational
function
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Any
polynomial with real coefficients can be factored uniquely into a
product of its lading coefficient, linear
factors
of the form (x
-
c)
that
correspond to its real roots, and
irreducible quadratic
polynomials that
correspond
to pairs of conjugate complex roots.
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Therefore,
to
represent a proper rational function as a finite sum of terms we
factor the polynomial Q
in the
denominator
into powers of distinct linear terms and/or powers of distinct
irreducible quadratic polynomials, i.e., which do not have real roots.
Thus,
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if the factors of
Q
in the denominator are
of the form (x
-
c)
then the partial fractions will be of the form
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where
c
is a real root of order k
and A1,
A2,
. . . , Ak
are unknown constants.
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If
the factors of Q
are
irreducible quadratic polynomials
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of
order k then,
the partial
fractions will be of the form
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where
M1,
M2,
. . . , Mk
and N1,
N2,
. . . , Nk
are unknown constants.
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Note
that the leading coefficient a
and the vertical translation y0
of a quadratic polynomial should be of the same
sign the quadratic to have pair of conjugate complex roots a
±
bi.
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Use
of the
partial fraction decomposition to integrate a proper rational
function, examples
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Unknown
constants A,
B,
and C
can be calculated
by plugging suitable values for x,
like roots for example,
into
given equation. Thus, by plugging the following values for x
into the equation
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x2
-
x + 3 = A(x
-
1)(x
+ 2) + B(x + 2) + C(x -
1)2
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for
x = 1
obtained is 3
= 3B,
B = 1
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for
x
= -
2
obtained is 9
= 9C,
C = 1
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for
x = 0
obtained is 3
= -
2A
+ 2B + C,
and after
substituting B
= 1 and C
= 1
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we get 3
= -
2A + 2 + 1,
A = 0.
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