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Unformatted text preview: B= .
2 and 5. I = 3
ln 2
4 6. I = 2
A(5 − x) + B (x − 1)
=
,
(x − 1)(5 − x)
(x − 1)(5 − x) x=5 3
dx .
+4 1. I = Explanation:
To ﬁnd A, B so that x=1 ex 5
4
ln
3
3 Explanation:
Set u = ex + 4. Then
du = ex dx = (u − 4) dx,
while
x=0 =⇒ u = 5, x = ln 2 =⇒ u = 6. 4
2 eakin (ace669) – PostClass 7.4 – morales – (56430)
In this case, 2 By partial fractions,
6 I=
5 1
A
Bx + C
=
+2
.
2 + 1)
(x + 1)(x
x+1
x +1 3
du .
u(u − 4) To evaluate this last integral we need to use
partial fractions: for then
1
1
1
1
.
=
−
u(u − 4)
4 u−4 u To determine A, B, and C multiply through
by (x + 1)(x2 + 1): for then
1 = A(x2 + 1) + (x + 1)(Bx + C )
= ( A + B ) x2 + ( B + C ) x + ( A + C ) , Thus
I= 6 3
4 5 3
ln u − 4 − ln u
4 = which after comparing coeﬃcients gives 1
1
du
−
u−4 u = 6
5 A = −B , .
5 I= 1
2 Consequently,
=
I= 3
4 ln 1
1
− ln
3
5 = 35
ln
.
43 1 1
.
2 1
x−1
dx
−2
x+1 x +1 0 1 1
2 0 1
dx −
x+1
1 + x2 0 003 A= Thus 6 u...
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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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