3/18/2014
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HW 6.3 (5507725)
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Question
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Question Details
SEssC alc2 6.3.019. [2560087]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
x2 + 1
(x − 5)(x − 4)2
dx
Solution or Explanation
x2 + 1
C
A
B
=
+
+
. Multiply both sides by (x − 5)(x − 4)2 to get
x−5
x−4
(x − 4)2
(x − 5)(x − 4)2
x2 + 1 = A(x − 4)2 + B(x − 5)(x − 4) + C(x − 5). Setting x = 4 gives 17 = −C ⇔ C = −17. Setting x = 5
gives 26 = A. Equating coefficients of x2 gives 1 = A + B, so B = −25. Thus,
x2 + 1
(x − 5)(x − 4)2
2.
17
26
25
−
−
x−5
x−4
(x − 4)2
dx =
dx = 26 ln|x − 5| − 25 ln|x − 4| +
Question Details
17
+ C.
x−4
SEssC alc2 6.3.012. [2165769]
-
Evaluate the integral.
1
0
x−4
x2 − 5x + 6
dx
Solution or Explanation
x−4
A
B
=
+
. Multiplying both sides by (x − 2)(x − 3) to get x − 4 = A(x − 3) + B(x − 2)
x−2
x−3
x2 − 5x + 6
x − 4 = Ax − 3A + Bx − 2B
x − 4 = (A + B)x + (−3A − 2B).
The coefficients of x must be equal and the constant terms are also equal, so A + B = 1 and −3A − 2B = −4. Adding twice
the first equation to the second gives us −A = −2
1
0
x−4
x2 − 5x + 6
1
dx =
0
2
1
−
x−2
x−3
⇔
A = 2, and hence, B = −1. Thus,
dx = 2 ln|x − 2| − ln|x − 3|
1
0
= (0 − ln 2) − (2 ln 2 − ln 3)
Another Method: Substituting 3 for x in the equation x − 4 = A(x − 3) + B(x − 2) gives −1 = B. Substituting 2 for x gives
−2 = −A
⇔
A = 2.
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Question Details
SEssC alc2 6.3.018. [2560015]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
5x2 + 2x − 5
x3 − x
dx
Solution or Explanation
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4.
Question Details
SEssC alc2 6.3.021. [2560280]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
x3 + 16
x2 + 16
dx
Solution or Explanation
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5.
Question Details
SEssC alc2 6.3.022. [2560189]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
x2 − x + 14
x3 + 7x
dx
Solution or Explanation
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6.
Question Details
SEssC alc2 6.3.025. [2560609]
-
Evaluate the integral. (Use C for the constant of integration.)
7x3 + x2 + 49x + 1
(x2 + 1)(x2 + 7)
dx
Solution or Explanation
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Question Details
SEssC alc2 6.3.024. [2560674]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
x2 − 6x − 17
dx
(x − 1)2(x2 + 1)
Solution or Explanation
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8.
Question Details
SEssC alc2 6.3.028. [2165396]
-
SEssC alc2 6.3.523.XP.MI. [2560711]
-
Evaluate the integral.
7
2
x
x2 + 6x + 13
dx
Solution or Explanation
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9.
Question Details
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
5r 2 dr
r +9
Solution or Explanation
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10.
Question Details
SEssC alc2 6.3.023.MI. [2560722]
-
Evaluate the integral. (Remember to use ln |u| where appropriate. Use C for the constant of integration.)
5
(x − 1)(x2 + 4)
dx
Solution or Explanation
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Question Details
SEssC alc2 6.3.035. [2378058]
-
Make a substitution to express the integrand as a rational function and then evaluate the integral.
64
x
x−9
16
dx
Let u =
x.
Solution or Explanation
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12.
Question Details
SEssC alc2 6.3.037. [2560866]
-
Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant
of integration.)
x3
9
x2 + 3
dx
Solution or Explanation
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13.
Question Details
SEssC alc2 6.3.038. [2165003]
-
Make a substitution to express the integrand as a rational function and then evaluate the integral.
3
x
x2 + x
1
dx
Solution or Explanation
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14.
Question Details
SEssC alc2 6.3.039. [2560142]
-
Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant
of integration.)
2e2x
e2x + 16ex + 63
dx
Solution or Explanation
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Assignment Details
Name (AID): HW 6.3 (5507725)
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