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Unformatted text preview: r comparing coeﬃcients gives
A = −B , C = 0, 1
A= .
4 16
.
x+4 Thus
I= x−4+ 16
x+4 dx . Consequently, Thus
I= 2 1
4 1 x
1
dx
−2
x x +4 I= 1
1
ln(x) − ln(x2 + 4)
4
2 = = x2
1
ln 2
8
x +4 2 with C an arbitrary constant. 1 006 2
1 x2
− 4x + 16 ln x + 4 + C
2 . Determine the integral Consequently, I=
I= 5
1
ln
8
2 10.0 points dx
.
(x − 1)(x + 3) .
1. I = 005 2. I = 1
ln
4 x−1
x+3 3. I = 1
ln
2 2x + 2
(x − 1)(x + 3) 10.0 points Determine the integral
I= 1
ln ((x − 1)(x + 3)) + C
2 x2
dx .
x+4 1. I = x2 − 4x + 4 ln x + 4 + C + C correct 4. I = ln ((x − 1)(x + 3)) + C +C eakin (ace669) – PostClass 7.4 – morales – (56430)
5. I = 1
ln
4 x+3
x−1 5. Area = 6 + +C Explanation:
By partial fractions,
1
A
B
=
+
,
(x − 1)(x + 3)
x−1 x+3 4 77
sq.units
ln
42 Explanation:
The graph of f is an even function opening
upwards, while the graph of g is a horizontal
straight line. Thus the required area is similar
to the shaded region in the ﬁgure below...
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This homework help was uploaded on 03/21/2014 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas at Austin.
 Spring '11
 STEPP
 Calculus

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