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Evaluate each indefinite integral.
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Integration by Substitution
8x 3
2
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5
4
(5 + ln (x))
3
−5
3
dx; −16x
u = −2x
(
−4x+45− 1) dx; u = −4x 4 4)
− 1 (5x 4 + 5) ⋅ 20x 3 dx; u = 5x 4 + 5
5)
dx; u = 5 + ln
5 2)
4
3
2
3
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Kuta
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Infinite
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(−3x 3 --+Infinite
(3x 4 + 4) 4 dx
9)
1) dx Calculus
10) 12x
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ration by Integration
Substitutionby Substitution
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Period____
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Period____
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Period____
Evaluate
each
Evaluate
each indefinite
indefinite
integral. Use
Use the
the provided
provided substitution.
substitution.
ate each indefinite
integral.
Use the integral.
provided
substitution.
oftware - Infinite Calculus
Name___________________________________
Kuta
Software
Infinite
Calculus
Name___________________________________
Evaluate
each --indefinite
integral.
8x 3
Kuta
Software
Infinite
Calculus
Name___________________________________
4
5
−5
4
5
5
3
4
4
5
−5
4
5
5
3
4
5
−5
3)
−
4
5− 1)) dx; u = −3x − 1
−15x
−4x
11))1 dx;
uu4== −4x
−− 11 u = −2x + 5
titution
Integrals
1)
−15x
−3x
−− 11 dx; u = −3x −2)1 −16x 3 (−4x 4 −2)
2)
−16x
−4x
dx;
−4x5 4 dx;
(−3x 5 for
Date________________
Period____
15x
−1)
1)Definite
dx;
u =((−3x
−3x
1) −16x
dx; u((=
−4x−4− −
(−2x + 5) Period____
50x
Substitution
for
Definite
3x
− 5Integrals
3 4 dx; Date________________
3 u = e 3x
4u = 5x 2 + 5
2 (e
3 by
Integration
Substitution
)
Date________________
Period____
3) 9)
6e 3x−9
cos
−
5
dx;
4)
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ss each definite
integral
terms
of u, but do not evaluate.
5xx2 +35x) + 4 dx
integral
in terms
of u,provided
but do not
evaluate.
Express
Evaluateeach
eachdefinite
indefinite
integral.
Use the
substitution.
1
2
5
+ ln8x( x))
3 1 4
2
3
3
5
4
Kuta Software - Infinite
Calculus
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dx; u 0(=5x4x
2) 4sec
−12x
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1= sec
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⋅ tan−Calculus
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dx;33 −uName_____________________________
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1
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−12
x
4
x
−
1
dx;
u
=
4
x
−
1
5
4
(4x x+ 1) 11)
(
)
−12x
−4x + 2 dx 5 x − 3
0
1) 20
xsin
2) 16
x − 2dx) dx; u = 4 x − 2
(3xx ⋅−sec
12)
3) (⋅ 415x
2 (5 x2 − 3) dx; u =
1)
0
Substitution
for
Definite
Integrals
−1 (4 x +
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Per
Evaluate
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SubstitutionPeriod____
for Definite Integrals
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each indefinite
integral.
Evaluate eachExpress
indefinite
integral.
each
definite
integral
terms ofintegral
u, but do
not evaluate.
e provided substitution.
Express
eachindefinite
in terms
of u, but do not evaluate.
3
2
3
9) −9 x (−3 x 0+ 1) dx
10) 12 x 3 (3 x 4 1+ 4) 4 dx
1
4
2 x (3x 4 + 3) 0dx4 2 8 x
3 (sec 3−3x
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(3x
5 ) dx
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33 2 + Kuta
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dx;
u
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4
x
+
1
4
x
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1
dx;
u =x 24(4xName________
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+
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8x
8x
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16x
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4x
−
2
dx;
u
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4x
−
2
Kuta
Software
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8x
44
1)
dx;
u
=
4
x
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1
2)
−12
x 3−
−1ln1)( x)dx; u
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2
3
4
3 4 5) 33
44
44 dx; u = 5
+
3)
−
dx;
u
=
−2x
+
5
3)
−
dx;
u
=
−2x
+
5
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2
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)
4
3
4
x
+
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20x
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by
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3
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11) −12 x (−4 x + 2) dx
(3 x 5 − 3) 5 ⋅ 15 x 4 dx
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Substitution
for
Integrals
Definite
Evaluate each indefinite
Use the integral.
provided
substitution.
1 integral.
Evaluate
each indefinite
Use the provided
substitution.
2
1
24x
2 2
6x3( (x 2 −4 1) )dx;
u = x 2 2− 41 2
4)
dx;
u
=
4x24
+xx4 1
5
4
4 2
each
definite
in terms of u, bu
(
6x 3x + 3 3)
dx;
u
+
x
4x
1Express
3=
xx3x
33x dx;
3 x u 8)
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)
2− 1) 2dx;4)u = 4x − 50
(
)
6)
4sec
4x
⋅
tan
4x
⋅
sec
4x
dx;
=
sec
6
x
−
1
u
=
x
−
1
dx;dx;
u=
4 x352x+2 +
4 integral
(
)
(
)
2
2
4x
+
4
3) 6e cos
e
−
5
dx;
u
=
e
−
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4)
u
2
4 1)
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) dx;
xsin (5 x −1)3) dx;
u = 5(x5 x − −3 3) dx;
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x=(55+xx4)+
20 xsin
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0
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x
+
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2
2
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(33x 5 − 3) 51 ⋅ 1524x 4x dx
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evaluate.
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1
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=
x
−
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dx;
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=
4 x 224
+ x4
+
5
4
(
)
3)
6
x
x
−
1
dx;
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=
x
−
1
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2
(
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2
x
csc
x
−
1
5
5 2
3
4
1((sec
(
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(
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5
5x
+
5
5
+
ln
x
2 3x 4 + 3
(
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4
x
+
4
(
))
(
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5
+
ln
x
4
−1
7)
36x
3x
+
3
dx;
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=
0
ln (3(xx−1
4x ⋅⋅ sec
5 + ln ( x)) 5)
)4x))
+ 4((4x
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4sec
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2
((4x
6)
4sec
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(4x4x) ⋅dx;
) )) dx;
)− dx;
4x ⋅ tan 4x
= sec
dx;
= 5 +each
(4x(-3xindefinite
−12x
1) dx;= 5u += ln
4x −) 1Name___________________________________
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xln
1
x
x
4
13) 0 (−2 x 4 − 4) for
⋅ −32
x 3 dx
x
(e 4x −−34)x5 ⋅⋅tan
Substitution
14)
8Date________________
e 4−3
dx
n for Definite
Integrals
3
4 Definite Integrals
4
Date________________
Period____
(
)
(
)
5)
−36
x
sec
3
x
+
3
⋅
tan
3
x
+
3
dx
6)
−9sec
x ⋅ sec 2 (sec −3 x) dx Period____
ate each definite
integral.
Evaluate each definite integral.
Express
each
integral
terms of u, but do not evaluate.
definite integral
terms4definite
of u, but
do notin
evaluate.
50
x
1
8) 0x(in
50 x
3 x− 1
3x
3x
4x − 1) dx;3)u = 64x
3x
3x
8x
cos (e -1−3)5) dx;
ucos
=16x
e (e 3−x 5− 5) dx;1 u =16
4)
dx; uLLC= 5 x 2 2+ 5 dx;
6
e
e
−
5
4)
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2
2
8
x
x
0 = 2x + 33
1 4x + 4 2
−
dx;
sec4 x(52 x+ 4+ 5)
sec (5 x + 5)
x(4x8+x 5) dx dx; u = 22x 2 + 3 1 6)
16)
2x
+33 dx;
dx 32u =
5)dx; u−
6) u =5x
2
2
2
3
2
3) 2
2
2
2
2
2
3
3
2
2
(2xdx;
) 215)
(
+u3=
4x
+
4
(
)
1)
dx;
u
=
4
x
+
1
−12
x
4
x
−
1
dx;
u
=
4
x
−
1
(
)
0(4x − 1) dx;2)u = 4x
3)
6
x
x
−
1
dx;
u
=
x
−
1
(
)
(
)
4x
+
1
2)
−12x
−
1
2
x
+
3
4
x
+
4
−3
0
2
2
(4 x 2 +−3x
1) ⋅ tan
1)
−1
6) −1 −9sec
0 −3x) dx
−3x ⋅ sec 2 (sec
0
1
each3 definite
integral.
4
4Evaluate
Evaluate
each definite integral. 4 x
5
13) (−2 x − 4) ⋅ −32 x dx
14) (e − 4) ⋅ 8e 4 x dx
0
1
1
1
8x 2 4 0
16 x
5
4
24x
3
4
2
8
x
16
5 dx;
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7)
36x
3
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8)
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1
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=
4x
−
1
2 xintegral.
3 (3x
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44 +
5)
−
dx;
u
=
2
x
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3
6)
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Evaluate
each
indefinite
Evaluate
each
indefinite
integral.
3
4
4
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4x
+
u =8)u2 x= x4x
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3x +2+ 33) dx;2 u = 23x5) + 3 −8) x(4x 2− dx;
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6x (3x + 3)7)dx;36x
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1
dx;
2
2
2
(
)
(
)
2
x
+
3
4
x
+
4
(
)
4x
+
4
−3
0
(
)
(
)
2
x
+
3
4
x
+
4
0
3
−3
0
5cos (−43 + ln 4 x )
4x
(
)
15)
x
4
x
+
5
dx
5
x
2
x
+
3
7)
−
dx
8)
dx
3
416)
4
3
4
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Evaluate
each
dx; u 2= 3 x
8)
u =definite
42 x 2 + 2integral.
24
x + 2 2 dx;
2
2
24x
2
( 4x )
(4xdx;+ u2)= 4)
u = x − 4)
1
0
4x 2 +0 4 (42x + 22 ) dx; u = 4 x + 4
2
2
−1
8)
dx
0 (4 x + 4) 0
0 (4x + 4)
8x
csc ( x 4 − 1)
5)
−
dx; u = 2 x 2 + 3
2
2
−3 (2 x + 3)
-2-16) 5 x 2 x +
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15) x(4 x + 5) 0dx
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1
8
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2
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18 x 2 (3 x 3 + 3)7) dx; 18
u =x 23(3x 3x 3++3 3) 2 dx; u = 3 x 3 + 38)
−
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18x (3x
2
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2
3
3
2
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3 x 3 +2 33
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x 2 −−116 x x −3 1 dx;
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