Supplement to 7.4
After talking to many of you in office hours, I decided to write this to clarify some
integration tricks that come up at the end of Partial Fractions problems.
Oftentimes, the
last step involves evaluating an integral of the form
2
ax
b
dx
cx
dx
e
+
+
+
∫
.
I will work
through a few examples to illustrate the principles involved.
The two identities to keep in mind are
ln
du
u
C
u
=
+
∫
and
(
)
1
1
2
2
tan
x
a
a
du
C
u
a
−
=
+
+
∫
Problem:
2
2
1
x
dx
x
+
+
∫
Solution:
Split the fractions into two parts.
(
)
(
)
2
2
2
2
1
1
2
2
1
2
1
1
ln
1
2tan
x
dx
x
x
dx
dx
x
x
x
x
C
−
+
=
+
+
=
+
+
+
+
+
∫
∫
∫
Make sure you see how to get each of the last two integrals.
Problem:
2
2
2
2
x
dx
x
x
+
+
+
∫
Solution:
Looking ahead to
2
2
2
w
x
x
=
+
+
, you would want the numerator to be
2
2
dw
x
=
+
.
We massage the fraction into two parts; one having the correct numerator
and the other being amenable to the
1
tan
−
identity above.
First, look at the
x
. We want that to be a
2
x
, so we multiply the numerator by 2 and
offset it with a fraction in the front.
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 Spring '10
 Williamson
 Division, Fractions, Fraction, Elementary arithmetic, Rational function, Mathematics in medieval Islam

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