Unformatted text preview: 4
3
ln
4
u B = βC , 1
0 x2 x
dx
+1 1
dx ,
+1 10.0 points
and so Evaluate the integral
1 I=
0 1. I = I= 1
dx .
(x + 1)(x2 + 1) Ο
1
ln(2) +
4
2 correct 1
1
ln(x + 1) β ln(x2 + 1) + tanβ1 (x)
2
2
= Consequently, Ο
1
ln(8) β
2. I =
2
2
3. I = 1
Ο
ln(8) β
4
2 1Ο
β ln(2)
4. I =
22
5. I = I= Explanation: Ο
1
ln(2) +
4
2 004 10.0 points Evaluate the integral
2 I= 1Ο
β ln(2)
42 Ο
1
ln(2) +
6. I =
2
2 (x + 1)2
1
ln 2
+ 2 tanβ1 (x)
4
x +1 1 1. I = 8
1
ln
4
5 2. I = 8
1
ln
16
5 1
dx .
x3 + 4 x . 1
0 . 1
0 eakin (ace669) β PostClass 7.4 β morales β (56430) 3 3. I = 1
5
ln
4
2 2. I = x2 β 4x + 16 ln x + 4 + C 4. I = 8
1
ln
8
5 3. I = x2 + 4x + 16 ln x + 4 + C 1
5
5. I = ln
correct
8
2 4. I = x2
+ 4x + 16 ln x + 4 + C
2 1
5
ln
16
2 5. I = x2
+ 4x + 4 ln x + 4 + C
2 6. I = x2
β 4x + 16 ln x + 4 + C correct
2 6. I = Explanation:
By partial fractions,
1
A Bx + C
=
+2
.
x3 + 4 x
x
x +4 Explanation:
After division we see that To determine A, B, and C multiply through
by x3 + 4x: for then x2 β 16
16
x2
=
+
x+4
x+4
x+4 1 = A(x2 + 4) + x(Bx + C ) = xβ4+ = (A + B )x2 + Cx + 4A ,
which afte...
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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.
 Spring '11
 STEPP
 Calculus

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