James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 7.6 Integration Using Tables and Computer Algebra Systems 511
Therefore
y xsx 2 1 2x 1 4 dx
− 1 3 sx 2 1 2x 1 4d 3y2 2 x 1 1
2
Computer Algebra Systems
sx 2 1 2x 1 4 2 3 2 lnsx 1 1 1 sx 2 1 2x 1 4 d 1 C
We have seen that the use of tables involves matching the form of the given integrand with
the forms of the integrands in the tables. Computers are particularly good at matching
pat terns. And just as we used substitutions in conjunction with tables, a CAS can perform
sub stitutions that transform a given integral into one that occurs in its stored formulas. So
it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean
that integration by hand is an obsolete skill. We will see that a hand computation sometimes
produces an indefinite integral in a form that is more convenient than a machine answer.
To begin, let’s see what happens when we ask a machine to integrate the relatively
simple function y − 1ys3x 2 2d. Using the substitution u − 3x 2 2, an easy calculation
by hand gives
y
1
3x 2 2 dx − 1 3 ln | 3x 2 2 | 1 C
whereas Mathematica and Maple both return the answer
1
3 lns3x 2 2d
The first thing to notice is that computer algebra systems omit the constant of integration.
In other words, they produce a particular antiderivative, not the most general one.
Therefore, when making use of a machine integration, we might have to add a constant.
Second, the absolute value signs are omitted in the machine answer. That is fine if
our prob lem is concerned only with values of x greater than 2 3 . But if we are interested in
other val ues of x, then we need to insert the absolute value symbol.
In the next example we reconsider the integral of Example 4, but this time we ask a
machine for the answer.
n
This is equation 3.11.3.
Example 5 Use a computer algebra system to find y xsx 2 1 2x 1 4 dx.
SOLUtion Maple responds with the answer
1
3 sx 2 1 2x 1 4d 3y2 2 1 4 s2x 1 2dsx 2 1 2x 1 4 2 3 s3
arcsinh
2 3
s1 1 xd
This looks different from the answer we found in Example 4, but it is equivalent
because the third term can be rewritten using the identity
Thus
arcsinh x − lnsx 1 sx 2 1 1d
arcsinh s3
3 s1 1 xd − ln F s3
3 s1 1 xd 1 s1 3 s1 1 xd2 1 1G
− ln
− ln
1
s3
f1 1 x 1 ss1 1 xd 2 1 3 g
1
s3
1 lnsx 1 1 1 sx 2 1 2x 1 4 d
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