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17.7 Rational Functions
In this section we present a few rules for finding antiderivatives of simple
rational
functions.
To antidifferentiate
where
is a polynomial,
make
the substitution
.
17.56
Example.
To find
![$\displaystyle { \int {{x^2+1}\over {(x-2)^2}}dx}$](img4328.gif)
.
Let
. Then
so
, and
To find
where
and
is a
polynomial of degree less than
.
We will find numbers
and
such that
![\begin{displaymath}
{{R(x)}\over {(x-a)(x-b)}}={A\over {(x-a)}}+{B\over {(x-b)}}.
\end{displaymath}](img4334.gif) |
(17.57) |
Suppose (17.57) were valid. If we multiply both sides by
we
get
Now take the limit as
goes to
to get
The reason I took a limit here, instead of saying ``now for
we get
'' is that
is not in the domain of the function we are
considering.
Similarly
and if we take the limit as
goes to
, we get
Thus,
![\begin{displaymath}
{{R(x)}\over {(x-a)(x-b)}}={1\over {a-b}}\Bigg[ {{R(a)}\over
{x-a}}-{{R(b)}\over {x-b}}\Bigg].
\end{displaymath}](img4341.gif) |
(17.58) |
I have now shown that if there are numbers
and
such that
(17.57) holds, then (17.58) holds. Since I have
not shown that such numbers exist, I will verify directly that
(17.58) is valid.
Write
. Then
17.59
Example.
To find
![$\displaystyle { \int {{x+1}\over {(x+2)(x+3)}}dx}$](img4344.gif)
.
Let
.
Then
so
and
so
Hence
In this example I did not use formula (
17.58), because I find it
easier to remember the procedure than the general formula. I do not
need to check my answer, because my proof of (
17.58) shows that
the procedure always works. (In practice, I usually do check the result
because I am likely to make an arithmetic error.)
To find
where
is a polynomial of
degree
, and
does not factor as a product of two first degree
polynomials.
Complete the square to write
Then
, since if
then we have factored
, and if
we
can write
, and then
and again we get a factorization of
. Since
, we can write
for
some
, and
Now
Make the substitution
to get an antiderivative of the form
The last antiderivative can be found by a trigonometric substitution.
17.60
Example.
To find
![$\displaystyle {\int {u\over {4u^2+8u+7}}du}$](img4365.gif)
:
Let
Let
![$t=u+1$](img4367.gif)
, so
![$u=t-1$](img4368.gif)
and
![$du=dt$](img4369.gif)
. Then
Now let
![$\displaystyle { t={{\sqrt 3}\over 2}\tan\theta}$](img4371.gif)
, so
![$\displaystyle { dt={{\sqrt 3}\over
2}\sec^2\theta \; d\theta}$](img4372.gif)
, and
![$\displaystyle { t^2+{3\over 4}={3\over 4}\sec^2\theta}$](img4373.gif)
.
Then
Hence,
To find
where
is a polynomial of
degree
.
First use long division to write
where
is a polynomial, and
is a polynomial of degree
. Then
use one
of the methods already discussed.
17.61
Example.
To find
![$\displaystyle {\int {{x^3+1}\over {x^2+1}}dx}$](img4379.gif)
.
By using long division, we get
Hence
17.62
Example.
In exercise
17.7, you showed that
![$\ln\Big(
\vert\sec(x)+\tan(x)\vert\Big)$](img4383.gif)
is an antiderivative for
![$\sec(x)$](img4384.gif)
. The function
![$\ln\Big(
\vert\sec(x)+\tan(x)\vert\Big)$](img4383.gif)
in that exercise appeared magically with no
motivation. I will now derive the formula, using standard methods:
Now let
![$u=\sin(x)$](img4386.gif)
. Then
![$du=\cos(x)\;dx$](img4387.gif)
, and
Suppose
![$\displaystyle { {1\over {1-u^2}}={A\over {1-u}}+{B\over {1+u}}}$](img4389.gif)
. Then
and if we take the limit of both sides as
![$u\to 1$](img4391.gif)
we get
![$\displaystyle {A={1\over 2}}$](img4392.gif)
.
Also
and if we take the limit as
![$u\to-1$](img4394.gif)
, we get
![$\displaystyle {B={1\over 2}}$](img4395.gif)
. Thus
Now
so
and thus
17.63
Exercise.
Criticize the following argument:
I want to find
. Suppose
Then
If we take the limit of both sides as
![$x\to 1$](img4403.gif)
, we get
![$\displaystyle { {1\over 2}=A}$](img4404.gif)
.
Also
and if we take the limit of both sides as
![$x\to -1$](img4406.gif)
, we get
![$\displaystyle { -{1\over
2}=B}$](img4407.gif)
.
Thus
Hence,
17.64
Exercise.
Find the following antiderivatives:
- a)
-
- b)
-
- c)
-
- d)
-
- e)
-
- f)
-
- g)
-
A
Next: Bibliography
Up: 17. Antidifferentiation Techniques
Previous: 17.6 Substitution in Integrals
  Index
Ray Mayer
2007-09-07