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S OLUTION. The strategy is not immediately obvious. There is no clear simpliﬁcation to be made, no trigonometric functions, no radicals, and no rational functions. We might be able to use integration by parts, using
13 WATERLOO SOS E XAM AID: MATH138 M IDTERM
the “1” trick.
Then
Notice is a rational function, so we can try using integration by parts. By our algorithm, we must ﬁrst
apply polynomial division since the degree of the numerator is at least the degree of the denominator. We
get
Notice the denominator is already factored into 1 term, so we have the partial fraction decomposition.
E XERCISE 2. Calculate
S OLUTION. This is a simple integration by parts:
so that
Using another integration by parts,
14 WATERLOO SOS E XAM AID: MATH138 M IDTERM
so that
E XERCISE 3. Calculate
in two ways, by making the substitution and by a suitable trigonometric substitution.
S OLUTION. Making the substitution , we get and . Then
Alternatively, we could make the trigonometric substitution so that . Then
Now we make the substitution so that .
To return to our original variable, we must construct the triangle generated by , yielding
. Then
As expected, our answers coincide. 15 WATERLOO SOS E XAM AID: MATH138 M IDTERM
E XERC...
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.
 Winter '07
 Anoymous
 Math, Calculus

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