

Integral
calculus


Integrating
rational functions

Recall
that a rational function is a ratio of two polynomials
written P_{n}(x)
/ Q_{m}(x). 
To
integrate a rational function



in
case n >
m, that is, if the
degree of P
is greater than or equal to degree of Q,
we first divide P_{n}(x)
by Q_{m}(x)
to obtain the sum of the polynomial P_{n
}_{}_{
m}(x)
and a proper or simple rational function.

Then,
we integrate proper rational function using decomposition of rational function
into a sum of partial fractions.


Use
of the
partial fraction decomposition to integrate a proper rational
function

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.

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,


if the factors in the denominator
Q
are
of the form (x

c)
then the partial fractions will be of the form


where
c
is a real root of order k
and A_{1},
A_{2},
. . . , A_{k}
are unknown constants.


If
the factors of Q
are
irreducible quadratic polynomials


of
order k then,
the partial
fractions will be of the form


where
M_{1},
M_{2},
. . . , M_{k}
and N_{1},
N_{2},
. . . , N_{k}
are unknown constants.

Note
that the leading coefficient a
and the vertical translation y_{0}
of a quadratic polynomial should be of the same
sign the quadratic to have pair of conjugate complex roots a
±
bi.


Use
of the
partial fraction decomposition to integrate a proper rational
function examples

Example:
Evaluate 




Example:
Evaluate 



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

x^{2}

x + 3 = A (x

1)(x
+ 2) + B (x + 2) + C (x 
1)^{2}

for
x = 1
obtained is 3
= 3B,
B = 1

for
x
= 
2 obtained is
9
= 9C,
C = 1

for
x = 0
obtained is 3
= 
2A
+ 2B + C,
and after
substituting B
= 1 and C
= 1

we get 3
= 
2A + 2 + 1,
A = 0.


Example:
Evaluate 


















Contents
M






Copyright
© 2004  2020, Nabla Ltd. All rights reserved.


