© Hong Kong Educational Publishing Co. 142
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
6
pp.214 – 227 p.214 2. xx +23 is a primitive function of .16 +x
3. 232 xx + is a primitive function of .26 2 xx +
4. xe2 is a primitive function of .2 2xe
p.216 (a) Cxdxx +=∫ 767
(b) Cxdxx
+=⎟⎠⎞
⎜⎝⎛−∫ 3
4 51
(c) Cxdxx +=∫ 321
52
15
(d) Cedte tt +=∫ 22
21
p.227 1. 13 −= xu
8)(
3
uuF
u
=
=′
2. 42 −= xu
uuF
xu
=
=′
)(
2
3. 4xu =
ueuF
xu
=
=′
)(
4 3
4. xxu += 2
6)(
12
uuF
xu
=
+=′
pp.217 – 240
6.1 (a) )ln2( Cxdxd
+ p.217
x
x2
12
=
⎟⎠⎞
⎜⎝⎛=
∴ The indefinite integral is correct.
(b) ⎥⎥⎦
⎤
⎢⎢⎣
⎡++ Cx
dxd 2
3
)2(21
2
43
)2(23
21
+=
+= •
x
x
∴ The indefinite integral is incorrect. 6.2 (a) Cxdx +=∫ 88 , where C is any constant p.220
(b) Cxdxx +=∫ 76
71 , where C is any constant
(c) Cxdxxdxx +== ∫∫ 34
31
3
43 ,
where C is any constant
(d) Cedxe xx +=∫ 66
61 , where C is any constant
6.3 (a) ∫ + dxxx )4( 3 p.222
212
4
21
2
2
1
4
3
4 where,24
constantsany are and where
,42
44
CCCCxxCC
CxCx
xdxdxx
+=++=
+⎟⎟⎠
⎞⎜⎜⎝
⎛++=
+= ∫ ∫
(b) ∫ +− − dxxx )236( 421
32132
3321
323
127
421
36 where,24
constantsany are and , where
,2_313
326
236
CCCCCxxx
CCC
CxCxCx
dxdxxdxx
+−=+++=
++⎟⎠⎞
⎜⎝⎛−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
+−=
−
−
−∫ ∫ ∫
Indefinite Integrals
143 © Hong Kong Educational Publishing Co.
Indefinite Integrals
6.4 (a) ∫ + dxxx )43( p.223
Cxx
Cxx
xdxdxx
xdxx
++=
+⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛=
+=
+=
∫∫∫
23
23
2
2
2
24
33
43
43
, where C is any constant
(b) ∫ −+ dxxx )13)(2( 2
Cxxxx
C
Cxxxx
dxxdxdxxdxx
dxxxx
+−+−=
+−⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛=
−+−=
−+−=
∫∫∫∫∫
2334
3
constantany is where
,22
634
3
263
)263(
23
4
234
23
23
(c) ∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛ + dxx
xx2
Cxx
dxx
++=
+= ∫
2
)1(2
, where C is any constant
(d) ∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛ +− dxx
xxx3
54 26
constantany is where,3
22
6
26
)26(
32
22
22
CCxxx
dxxxdxdxx
dxxxx
++−−=
+−=
+−=
∫∫∫∫
−
−
(e) ∫ + dxxe x )65( 22
constantany is where,2
25
65
32
22
CCxe
dxxdxe
x
x
++=
+= ∫∫
(f) dxx
xex
∫−1
Cxe
dxxex
x
+−=
−= ∫ −
ln
1
, where C is any constant
6.5 (a) )( xxedxd p.224
xx exe +=
(b) ,)( 1Cxedxexe xxx +=+∫ 1C is any constant
,Cxedxedxxe xxx +=+∫∫
,12 CxeCedxxe xxx +=++∫ 2C is any constant
∫ +−= ,Cexedxxe xxx where 21 CCC −=
6.6 Let ,13 −= xu ,3=′u .)( 8uuF = p.228
∫ − dxx 8)13(3
Cx
CCu
duu
dxuu
+−
=
+=
=
′=
∫∫
9)13(
constantany is where,91
9
9
8
8
6.7 Let .42 −= xu p.229 xu 2=′
∴ duudxxn ∫∫ =− 42 2
Cx
C
Cu
+−=
+=
23
2
23
)4(32
constantany is where
,32
6.8 Let .13 −= xu p.230 ,3=′u dxdxu 3=′
dxxdxx ∫∫ −− −=− )3()13(31)13( 22
Cx
CCu
duu
+−
−=
+−=
=
−
−
−∫
3)13(
constantany is where,31
31
1
1
2
6.9 (a) Let .83 −= tu p.231 ,3 2tu =′ dttdtu 23=′
dtttdttt )3(318282 3232 ⎟⎠⎞
⎜⎝⎛−=− ∫∫
Ct
C
Cu
duu
+−=
+=
×=
•
∫
23
3
23
)8(94
constantany is where
,32
32
312
© Hong Kong Educational Publishing Co. 144
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
(b) Let .2xu −= ,2xu −=′ xdxdxu 2−=′
duedxxe ux ∫∫ −=−
212
Ce
CCe
x
u
+−=
+−=
−
2
constantany is where,21
2
(c) Let .23 −= xu ,3 2xu =′ dxxdxu 23=′
∫∫ =− u
dudxx
x31
23
2
C
x
CCu
+−
=
+=
3
2ln
constantany is where,ln31
3
6.10 Method I p.232 Let .1+= xu ,1=′u dudxu =′ and .1−= ux
duuudxxx ∫∫ −=+ )1(1
Cxx
CCuu
duuu
++−+=
+−=
−= ∫
23
25
23
25
21
23
)1(32)1(
52
constantany is where,32
52
)(
Method II Let ,1+= xu then ,12 += xu ,1 2ux −= .12 =′− uu
∴ dxx
dxudu12
1+
−=′=
dxxxxdxxx ∫∫ +
+−=+
12)1(21
Cxx
CCuu
duuu
duuu
++−+=
+−=
−=
−−=
∫∫
23
25
35
24
22
)1(32)1(
52
constantany is where,3
252
)22(
)1(2
6.11 ∫ + dxex x )1(2
p.233
∫ ∫+= dxexxdx x2
For the second integral, let ,2xu = ,2xu =′ .2xdxdxu =′
∴ ∫∫ = duedxxe ux
212
Ceu +=21 , where C is any constant
∴ ∫ ∫∫ +=+ duexdxdxex ux
21)1(
2
Cex x++=
22
22
6.12 ∫ −= dxey x 12 p.236
Cxex +−= 2 , where C is any constant
Q The curve passes through (0, 6). ∴ Ce +−= 026 0 4=C ∴ The equation is .42 +−= xey x
6.13 ∫ += xdxxdxdy 52 p.237
Cxx++=
25
3
23, where C is any constant
When x = 0, .0=dxdy
∴ C++= 000 ∴ 0=C
∴ 2
53
23 xxdxdy
+=
constantany is where,6
512
25
3
1
34
23
CCxx
dxxxy
++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+= ∫
When x = 0, y = 1.
1
001
1
1
=++=
CC
∴ The equation of the curve is
.16
512
34++=
xxy
145 © Hong Kong Educational Publishing Co.
Indefinite Integrals
6.14 (a) dtt
tx ∫ +−=
41202 p.238
dtt
t∫ +
−=4
120 2
Let .42 += tu ,2tu =′ tdtdtu 2=′
∴ ∫−
= dtu
x 12120
Ct
CCu
++−=
+−=
)4ln(60
constantany is where,||ln602
When t = 0, x = 300.
∴ C+−= 4ln60300 4ln60300 +=C ∴ 4ln60300)4ln(60 2 +++−= tx
⎟⎠⎞
⎜⎝⎛
++=
44ln60300 2t
(b) ⎟⎠⎞
⎜⎝⎛
++=
4104ln60300 2x
105= (cor. to the nearest integer) ∴ The concentration is 105 .unit/m3
6.15 (a) dtetw t∫ −= 25.06)( p.240
Ce
C
Ce
t
t
+−=
+−
=
−
−
25.0
205.0
24
constantany is where
,25.0
6
When t = 1, w = 1.
25.0
25.0
241
241−
−
+=
+−=
eC
Ce
∴ 25.025.0 24124 −− ++−= eew t )(241 25.025.0 tee −− −+=
(b) When t = 3,
3544.8
)(241 75.025.0
≈−+= −− eew
When t = 12,
4963.18
)(241 325.0
≈−+= −− eew
∴ The increase in weight kg )3544.84963.18( − kg 10= (cor. to the nearest kg)
pp.217 – 239 Example 6.1T p.217
(a) ⎟⎠⎞
⎜⎝⎛ +− C
xdxd 3
2
2
3
)](3[
x
xdxd
=
−−= −
∴ The indefinite integral is correct.
(b) ⎟⎟⎠
⎞⎜⎜⎝
⎛+C
xdxd
34
23
23
21
3
2
21
34
34
x
x
Cxdxd
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
−
−
∴ The indefinite integral is incorrect. Example 6.2T p.220
(a) Cxdx +−=−∫ 6)6( , where C is any constant
(b) ∫∫ = dxxdxx 21
Cx += 23
32 , where C is any constant
(c) Cedxe xx +=∫ 33
31 , where C is any constant
Example 6.3T p.222
dxxx )64( 5∫ −
2123
6
21223
1
6
21
5
64 where,432
constants are and where,326
64
64
CCCCxx
CCCxCx
dxxdxx
−=+−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+−⎟
⎟⎠
⎞⎜⎜⎝
⎛+=
−= ∫∫
© Hong Kong Educational Publishing Co. 146
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
Example 6.4T p.222
(a) dxex
x∫ ⎟⎠⎞
⎜⎝⎛ − −374
constantany is where,
37||ln4
714
3
3
CCex
dxedxx
x
x
++=
−=
−
−∫∫
(b) dxx
xx∫ ⎟⎟
⎠
⎞⎜⎜⎝
⎛ −3
32 54
constantany is where,5||ln4
)54( 1
CCxx
dxx
+−=
−= ∫ −
(c) xdxx∫ +− )12)(23(
constantany is where,2
22
)26(2
3
2
CCxxx
dxxx
+−−=
−−= ∫
Example 6.5T p.224
(a) )ln( 4 xxdxd
xxx
xxx
x
ln4
ln41
33
34
+=
+⎟⎠⎞
⎜⎝⎛=
(b) 1
433 ln)ln4( Cxxdxxxx∫ +=+ ,
where 1C is any constant
∫
∫
∫∫
+−=
+=++
+=+
,4
lnln4
constantany is where
,lnln44
lnln4
3
443
2
143
2
4
1433
Cxxxxdxx
C
CxxxdxxCx
Cxxxdxxdxx
where 213 CCC −=
∴ ,16
ln41ln
443∫ +−= Cxxxxdxx
where4
3CC =
Example 6.6T p.228
(a) Let .53 += xu ,3 2xu =′ dxxdxu 23=′
Cx
C
Cu
duudxxx
++
=
+=
=+ ∫∫
7)5(
constantany is where
,7
)3()5(
73
7
6263
(b) Let .2xu = ,2xu =′ .2xdxdxu =′
Ce
CCe
duedxxe
x
u
ux
+=
+=
= ∫∫
2
2
constantany is where,
2
Example 6.7T p.229
Let .13 += xu ,3 2xu =′ dxxdxu 23=′
Cx
CCu
duu
duudxxx
++=
+=
=
=+
•
∫
∫∫
23
3
23
21
32
)1(92
constantany is where,32
31
31
311
Example 6.8T p.229
Let .21 xu += ,2=′u dxdxu 2=′
Cx
CCu
duu
duu
dxxdxx
++=
+=
=
=
⎟⎠⎞
⎜⎝⎛+=+
•
∫
∫
∫∫
34
34
31
3
33
)21(83
constantany is where,43
21
2121
)2(212121
147 © Hong Kong Educational Publishing Co.
Indefinite Integrals
Example 6.9T p.230
(a) Let .54 −= xu ,4 3xu =′ dxxdxu 34=′
Cx
C
Cu
duu
dxxxdxxx
+−
=
+=
=
⎟⎠⎞
⎜⎝⎛−=−
•
∫
∫∫
32)4(
constantany is where
,81
4141
)4(41)4()4(
84
8
7
743743
(b) Let .23 += xu ,3 2xu =′ dxxdxu 23=′
Ce
CCe
due
dxexdxex
x
u
u
xx
+=
+=
=
⎟⎠⎞
⎜⎝⎛=
+
++
∫
∫∫
3
constantany is where,3131
)3(31
2
2222
3
33
(c) Let .8−= tu ,1=′u dtdtu =′
CtCCu
duu
dtt
dtx
+−=+=
=
−=
−
∫
∫∫
|8|ln4constantany is where,||ln4
14
814
84
Example 6.10T p.231
Let .43 += xu
,3=′u dxdxu 3=′ and )4(31
−= ux
Cxx
C
Cuu
duuu
duu
u
duu
udx
xx
++−+=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−=
−=
−=
−=
+
•
−
•
∫
∫
∫∫
21
23
21
23
21
21
)43(9
16)43(274
constantany is where
,2432
92
)4(92
492
)4(31
312
432
Example 6.11T p.233
∫ ++ dxxx ])1(1[ 42
∫∫∫
++=
++=
dxxxxdx
dxxxx42
42
)1(
])1([
For the second integral, let .12 += xu
,2xu =′ xdxdxu 2=′
Then ∫ ∫=+ dxudxxx 442
21)1(
∴ ∫ ++ dxxx ])1(1[ 42
constantsany are and where
,51
21
2
21
21
25
1
2
4
CC
CuCx
duuxdx
+++=
+=
•
∫∫
Cxx+
++=
10)1(
2
522
, where .21 CCC +=
Example 6.12T p.235
constantany is where,32
2
)2(
34
23
23
CCxx
duxdxx
dxxxy
++=
+=
+=
∫∫∫
Q The curve passes through (1, 2).
∴ 631
212 ++=
67
=C
∴ The equation of the curve is
.67
32
34++=
xxy
Example 6.13T p.236
constantany is where,26
)212(
2 CCxx
dxxdxdy
+−=
−= ∫
When x = 1, .0=dxdy
4260
−=+−=
CC
∴ 426 2 −−= xxdxdy
constantany is where,42
)426(
1123
2
CCxxx
dxxxy
+−−=
−−= ∫
© Hong Kong Educational Publishing Co. 148
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
When x = 1, y = 4.
7
4124
1
1
=+−−=
CC
∴ The equation of the curve is
.742 23 +−−= xxxy
Example 6.14T p.237
(a) dtt
h ∫ += 2)4(
1
Let 4+= tu .
,1=′u dtdtu =′
Ct
CCu
duu
duu
h
++
=
+−=
=
=
∫
∫−
41
constantany is where,1
1
2
2
When h = 6, t = 8.
1273
1216
4816
=
+−=
++
−=
C
C
C
∴ 1273
41
++
−=t
h
(b) When t = 0,
635
1273
41
=
+−=h
∴ The height is m. 6
35
Example 6.15T p.239
At
AAt
dtttC
+=
+=
=
•
∫
34
34
31
49
constantany is where,433
3)(
When t = 0, C = 40.
40
)0(4940
=
+=
A
A
∴ 4049)( 3
4
+= ttC
When t = 3,
40)3(49 3
4
+=C
500= (cor. to the nearest ten thousand)
∴ The cost is 500 thousand dollars.
6.1 pp.224– 226
p.224
1. (a) ∫ ++=+ Cxxdxx 2)12( ,
where C is any constant
(b) ∫ ++=− −− Cexdxe xx 2)2( ,
where C is any constant
(c) ∫ ++=+ Cexdxex xx 2342 23)49( ,
where C is any constant
(d) ∫ +=⎟⎟⎠
⎞⎜⎜⎝
⎛ − Cx
edxx
exe xxx
2 ,
where C is any constant
2. (a) ⎟⎟⎠
⎞⎜⎜⎝
⎛++ Cxx
dxd 2
3
3
)2(
2
2)3(31
2
2
+=+=
+=
xxxx
xx
∴ The indefinite integral is correct.
(b) ⎥⎦⎤
⎢⎣⎡ ++ Cx
dxd 82 )1(
161
72
72
)1(
)2()1(8161
+=
+= •
xx
xx
∴ The indefinite integral is correct. 3. ∫ += Cxdx 66 , where C is any constant
4. ∫ += Cxdxx5
54 , where C is any constant
5. ∫ += Cedxex
x
2
22 , where C is any constant
6. ∫ +−=−
− Ctdxt2
23 , where C is any constant
149 © Hong Kong Educational Publishing Co.
Indefinite Integrals
7. ∫ ∫= dxxdxx 53
5 3
Cx += 58
85 , where C is any constant
8. ∫ ∫−
= dxxdxx
53
5 3
1
Cx += 52
25 , where C is any constant
9. ∫ ∫ −− = dttdtt 77 66
16
116
6 ,
constantany is where,6616
CCCt
CCt
=+−=
+⎟⎠⎞
⎜⎝⎛−=
−
−
10. ∫ ∫= dxx
dxx
144
1
11
4 , ||ln4constantany is where,4 ||ln4
CCCxCCx
=+=+=
11. ∫ ∫ −− = dxedxe xx 22 33
1
2
112
3 ,23
constantany is where,213
CCCe
CCe
x
x
=+−=
⎟⎠⎞
⎜⎝⎛ +−=
−
−
12.
137
1137
34
5 ,7
15
constantany is where,735
5
CCCx
CCx
dxx
=+=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
= ∫
13. ∫ + dxxx )6( 34
214
521
2
4
1
5
34
6 ,23
5
constantsany are and where
,4
65
6
CCCCxx
CC
CxCx
dxxdxx
+=++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
+= ∫ ∫
14. ∫ −+ − dxxx )423( 22
32113
321
321
1
3
22
23 ,42
constantsany are and , where
,4)(23
3
423
CCCCCxxx
CCC
CxCxCx
dxdxxdxx
++=+−−=
+−+−+⎟⎟⎠
⎞⎜⎜⎝
⎛+=
−+=
−
−
−∫ ∫ ∫
15. ∫ + dxx
x )52(3 5
4 3
2132
47
21
232
147
35
43
52 ,7
1578
constantsany are and where
,235
742
53
CCCCxx
CC
CxCx
dxxdxx
+=+−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+−+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
+=
−
−
−
∫ ∫
16. ∫ ⎟⎠⎞
⎜⎝⎛ − − dxe
xx234
21
221
22
1
2
34 ,2
3||ln4
constantsany are and where
,213) ||(ln4
314
CCCCex
CC
CeCx
dxedxx
x
x
x
−=++=
⎟⎠⎞
⎜⎝⎛ +−−+=
−=
−
−
−∫ ∫
17. ∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛+ dx
xx 32
2121
23
21
221
123
21
21
32 ,634
constantsany are and where
,23322
32
CCCCxx
CC
CxCx
dxxdxx
+=++=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛++
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
+= ∫ ∫−
18. ∫ + dxxx )6(3
Cxx
CCxx
dxxdxx
++=
++⎟⎟⎠
⎞⎜⎜⎝
⎛=
+= ∫ ∫
23
23
2
9
constantany is where,2
183
3
183
19. ∫ +− dxee xx )2(
constantany is where,2
2
)2( 0
CCex
dxedx
dxee
x
x
x
++=
+=
+=
∫∫∫
20. ∫ ++ dxxx )1)(6(
constantany is where,627
3
67
)67(
23
2
2
CCxxx
dxxdxdxx
dxxx
+++=
++=
++=
∫ ∫∫∫
© Hong Kong Educational Publishing Co. 150
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
21. ∫ +− dxxx )2)(32(
constantany is where,623
2
62
)62(
23
2
2
CCxxx
dxxdxdxx
dxxx
+−+=
−+=
−+=
∫∫∫∫
22. ∫ ++ dxxx )2)(1( 2
constantany is where,223
24
22
)22(
23
4
23
23
CCxxxx
dxxdxdxxdxx
dxxxx
++++=
+++=
+++=
∫∫∫∫∫
23. ∫ ⎟⎠⎞
⎜⎝⎛ + dx
x
2
11
constantany is where,||ln2
21
121
1
2
2
CCxxx
dxdxx
dxx
dxxx
+++−=
++=
⎟⎠⎞
⎜⎝⎛ ++=
−
∫∫∫
∫
24. ∫+ dxx
x1
constantany is where,||ln
11
CCxx
dxx
dx
dxx
++=
+=
⎟⎠⎞
⎜⎝⎛ +=
∫∫
∫
25. ∫−+ dx
xxx
2
2 43
constantany is where,4 ||ln3
43
431
1
2
2
CCxxx
dxx
dxx
dx
dxxx
+++=
++=
⎟⎠⎞
⎜⎝⎛ −+=
−
∫∫∫
∫
p.226
26. ∫+ dxxx 2)1(
constantany is where,||ln22
12
12
2
2
CCxxx
dxx
x
dxx
xx
+++=
⎟⎠⎞
⎜⎝⎛ ++=
++=
∫
∫
27. ∫+ dtt
t2
22 )1(
constantany is where,123
12
12
3
22
2
24
CCt
tt
dtt
t
dtt
tt
+−+=
⎟⎠⎞
⎜⎝⎛ ++=
++=
∫
∫
28. ∫ + dxex 2)2(
Cxee
CCxee
dxee
xx
xx
xx
+++=
++⎟⎟⎠
⎞⎜⎜⎝
⎛+=
++= ∫
422
constantany is where,42
42
)44(
22
22
2
29. ∫− dx
ee
x
x 13 2
Cee
CCee
dxee
xx
xx
xx
++=
+−−=
−=
−
−
−∫
3
constantany is where,)(3
)3(
30. ∫−−+ dx
eeee
x
xxx
22 3
Cexe
Cdxee
xx
xx
+++=
⎟⎠⎞
⎜⎝⎛ ++=
−
−∫
422
constantany is where,21
212
22
22
31. ∫+ dxe
ex
x 2)2(
constantany is where,44
)44(
442
CCexe
dxee
dxe
ee
xx
xx
x
xx
+−+=
++=
++=
−
−∫
∫
32. ∫+ dxxe
xeex
xx
32 3
Cex
CCex
dxex
x
x
x
++=
++=
⎟⎠⎞
⎜⎝⎛ +=
•
∫
2
2
2
65||ln
32
constantany is where,21
35||ln
32
35
32
151 © Hong Kong Educational Publishing Co.
Indefinite Integrals
33. ∫ −− dx
xx
242
constantany is where,22
)2(2
)2)(2(
2CCxx
dxx
dxx
xx
++=
+=
−+−
=
∫∫
34. ∫ ++ dx
xx
113
constantany is where,23
)1(1
)1)(1(
25
2
2
CCxxx
dxxx
dxx
xxx
++−=
+−=
++−+
=
∫∫
35. (a) )( 3xxedxd
xx xee 33 3+=
(b) 1333 )3( Cxedxxee xxx +=+∫ ,
where 1C is any constant
∫∫ +=+ 1333 3 Cxedxxedxe xxx
133
2
33
3CxedxxeCe xx
x+=++ ∫ ,
where 2C is any constant
21
333
33 CCexedxxe
xxx −+−=∫
3
,93
21
333
CCC
Cexedxxexx
x
−=
+−=∫
36. (a) )ln( 5 xxdxd
xxx
xxx
x
ln5
ln51
44
45
+=
+⎟⎠⎞
⎜⎝⎛=
(b) 1544 ln)ln5( Cxxdxxxx +=+∫ ,
where 1C is any constant
∫∫ +=+ 1544 lnln5 Cxxxdxxdxx
154
2
5lnln5
5CxxxdxxCx
+=++ ∫ ,
where 2C is any constant
5
,255
lnln
21
554
CCC
Cxxxxdxx
−=
+−=∫
37. (a) xdxd ln
xx
xdxd
ln21
])[(ln 21
=
=
(b)
constantany is where
,lnln2
1
1
1
C
Cxdxxx
+=∫
1lnln1
21 Cxdx
xx+=∫
12,ln2ln1 CCCxdx
xx=+=∫
38. (a) Let 42)( xxg =′ .
By Definition 6.2, ∫ += Cdxxxg 42)( ,
where C is any constant
( ) 44 2)(2 xxgdxxdxd
=′=∴ ∫
(b) Let xxxg ln)( =′ .
By Definition 6.2, ∫ += Cxdxxxg ln)( ,
where C is any constant
∴
( )
xx
xgxdxxdxd
ln
)(ln
=
′=∫
39. (a) Let ).()( 2 xedxdxf x += Then xexg x += 2)( .
By Definition 6.2,
dxxfdxxedxd x ∫∫ =+ )()( 2
Cxe x ++= 2 ,
where C is any constant
(b) Let ).ln()( 4 xxdxdxf = Then xxxg ln)( 4= .
By Definition 6.2,
,ln
)()ln(
4
4
Cxx
dxxfdxxxdxd
+=
= ∫∫
where C is any constant 40. (a) Let 3)( xxg =′ .
By Definition 6.2, ∫= dxxxg 3)( .
( )
3
3
)(
x
xgdxxdxd
=
′=∴ ∫
© Hong Kong Educational Publishing Co. 152
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
(b) Let )()( 3xdxdxf = . Then 3)( xxg = .
By Definition 6.2,
constantany is where,
)(
3
3
CCx
dxxfdxxdxd
+=
=⎟⎠⎞
⎜⎝⎛
∫∫
(c) No, they are not equal.
( )∫ dxxdxd 3 is a function, but
dxxdxd
∫ ⎟⎠⎞
⎜⎝⎛ 3 is a group of functions.
6.2 pp.233 – 235
p.233
1. Let .72 −= xu ,2xu =′ .2xdxdxu =′
∫∫ =− duudxxx 882 )7(2
Cx
CCu
+−
=
+=
9)7(
constantany is where,9
92
9
2. Let .14 += xu ,4=′u .4dxdxu =′
∫∫ =+ duudxx 1010)14(4
Cx
CCu
++
=
+=
11)14(
constantany is where,11
11
11
3. Let .23 += tu ,3 2tu =′ .3 2dttdtu =′
∫∫ =+ duudttt 23 32
Ct
CCu
++=
+=
23
3
23
)2(32
constantany is where,32
4. Let .12 += eu ,xeu =′ .dxedxu x=′
∫∫ =+ duudxee xx 44)1(
Ce
CCu
x+
+=
+=
5)1(
constantany is where,5
5
5
5. Let .2xu = ,2xu =′ .2xdxdxu =′
Ce
CCe
duedxxe
x
u
ux
+=
+=
= ∫∫
2
2
constantany is where,
2
6. Let .14 2 += xu ,8xu =′ .8xdxdxu =′
Cx
CCu
duudxxx
++
=
+=
=+ ∫∫
7)14(
constantany is where,7
)14(8
72
7
662
7. Let .42 += tu ,2tu =′ .2tdtdtu =′
∫∫ =+
duu
dtt
t32
1)4(
2
Ct
CCu
duu
++
−=
+−=
=
−
−
−∫
2)4(
constantany is where,2
22
2
3
8. Let .32 ttu += ,32 +=′ tu .)32( dttdtu +=′
∫∫ =++ duudtttt 3)32( 2
Ctt
C
Cu
++=
+=
23
2
23
)3(32
constantany is where
,32
9. Let .13 += xu ,3=′u .3dxdxu =′
Cx
CCu
duudxx
++
=
+=
=+
•
∫∫
21)13(
constantany is where,73
131)13(
7
7
66
10. Let .12 2 += xu ,4xu =′ .4xdxdxu =′
Cx
CCu
duuxdxx
++
=
+=
=+
•
∫∫
24)12(
constantany is where,64
141)12(
62
6
552
153 © Hong Kong Educational Publishing Co.
Indefinite Integrals
11. Let .52 −= tu ,2=′u .2dtdtu =′
Ct
CCu
duudtt
+−
−=
+−=
=−
−
−•
−− ∫∫
2)52(
constantany is where,2121)52(
1
1
22
12. Let .1 3xu += ,3 2xu =′ .3 2dxxdxu =′
Cx
CCu
duu
dxx
x
++
=
+=
=+ ∫∫
3|1|ln
constantany is where,||ln31
131
1
3
3
2
p.234
13. Let .64 −= xu ,4 3xu =′ .4 3dxxdxu =′
∫∫ =− duudxxx41643
Cx
CCu
+−
=
+= •
6)6(
constantany is where,32
41
23
4
23
14. Let .843 2 ++= xxu ,46 +=′ xu .)46( dxxdxu +=′
Cxx
CCu
duu
dxxx
x
+++=
+=
=++
+∫∫
21
2
21
2
)843(2
constantany is where,2
1
843
46
15. ∫ •⎟⎟⎠
⎞⎜⎜⎝
⎛ + dxxx
x3
2
2
2 12
∫ •⎟⎠⎞
⎜⎝⎛ += dx
xx 3
2
2121
Let .21 2xu +=
,43x
u −=′ .43 dx
xdxu −=′
∫∫ −=⎟⎠⎞
⎜⎝⎛ + • duudx
xx2
3
2
2 41121
Cx
C
Cu
+⎟⎠⎞
⎜⎝⎛ +−=
+−= •
3
2
3
21121
constantany is where
,34
1
16. Let .52 += teu ,2 2teu =′ .2 2 dxedxu x=′
Ce
CCu
duudxee
x
xx
++
=
+=
=+
•
∫∫
10)5(
constantany is where,52
121)5(
52
5
4422
17. Let .22 teu t −= ),1(222 22 −=−=′ tt eeu .)1(2 2 dtedtu t −=′
∫∫ =−− duudtete tt 2222
21)1()2(
Cte
C
Cu
t+
−=
+= •
6)2(
constantany is where
,32
1
32
3
18. Let .92 2 += tu ,4tu =′ .4tdtdtu =′
Ct
CCu
duu
dtt
t
++
−=
+−
=
=+
−
−•
∫∫
12)92(
constantany is where,3
141
141
)92(
32
3
442
19. Let .ln tu =
,1t
u =′ .1 dtt
dtu =′
Ct
CCu
duudttt
+=
+=
=
•
∫∫
8)(ln
constantany is where,42
121
2)(ln
4
4
33
© Hong Kong Educational Publishing Co. 154
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
20. Let .baxu += ,au =′ .adxdxu =′
Ca
e
CCea
duea
dxe
bax
u
ubax
+=
+=
=
+
+
•
∫∫
constantany is where,1
1
21. Let .12 += xu ,2xu =′ .2xdxdxu =′
Ce
CCe
duexdxe
x
u
ux
+=
+=
−=
+
+ ∫∫
2
constantany is where,21
21
1
1
2
2
22. Let .3 2xu = ,6xu =′ .6xdxdxu =′
Ce
CCe
duedxxe
x
u
ux
+=
+=
= ∫∫
3
constantany is where,31312
2
2
3
3
23. Let .1t
u =
,12t
u −=′ .12 dt
tdtu −=′
Ce
CCe
duedtet
t
u
ut
+−=
+−=
−= ∫∫
1
1
2
constantany is where,
1
24. Let .32 xxu −= ,32 −=′ xu .)32( dxxdxu −=′
Ce
CCe
duedxex
xx
u
uxx
+=
+=
=−
−
− ∫∫
3
3
2
2
constantany is where,
)32(
25. Let .33 xxu += ,33 2 +=′ xu .)1(3 2 dxxdxu +=′
constantany is where
,
)1(33
43 2
3
3
C
Ce
duedxexxx
xx
+=
=+
+
+ ∫∫
26. Let .12 ++= ttu ,12 +=′ tu .)12( dttdtu +=′
Ce
CCe
duedtet
tt
u
utt
+=
+=
=+
++
++ ∫∫
1
1
2
2
constantany is where,
)12(
27. Let .xu =
,2
1x
u =′ .2
1 dxx
dxu =′
Ce
CCe
duedxx
e
x
u
ux
+=
+=
= ∫∫
2
constantany is where,2
2
28. Let .82 += xu ,2xu =′ .2xdxdxu =′
Cx
CCu
duu
dxx
x
++=
+=
=+ ∫∫
|8|ln21
constantany is where,||ln21
121
8
2
2
29. Let .1 2xu −= ,2xu −=′ .2xdxdxu −=′
Cx
CCu
duu
dxxx
+−
−=
+−=
⎟⎠⎞
⎜⎝⎛−=
− ∫∫ •
2|1|ln3
constantany is where,||ln23
1213
13
2
2
30. Let .1 xu −= ,1−=′u .dxdxu −=′
Cx
CCu
duu
dxx
+−−=+−=
−=− ∫∫
|1|ln2constantany is where,||ln2
121
2
31. Let .22 ttu += ,22 +=′ tu .)1(2 dttdtu +=′
Ctt
CCu
duu
dttt
t
++=
+=
=++
∫∫
|2|ln21
constantany is where,||ln21
121
21
2
2
155 © Hong Kong Educational Publishing Co.
Indefinite Integrals
32. Let .12 += xeu ,2 2xeu =′ .2 2 dxedxu x=′
Ce
CCu
duu
dxe
e
x
x
x
++=
+=
=+ ∫∫
|1|ln21
constantany is where,||ln21
121
1
2
2
2
33. Let .23 xxu += ,23 2 +=′ xu .)23( 2 dxxdxu +=′
Cxx
CCu
duu
dxxx
xdxxx
x
++
=
+=
=
++
=++
∫
∫∫
5|2|ln
constantany is where,||ln51
151
)2(523
10523
3
3
2
3
2
34. Let .122 24 +−= xxu ),4(228 33 xxxxu −=−=′ .)4(2 3 dxxxdxu −=′
Cxx
CCu
duu
dxxx
xx
++−
=
+=
=+−
−∫∫
2|122|ln
constantany is where,||ln51
121
1224
4
24
3
35. Let .ln xu =
,1x
u =′ .1 dxx
dxu =′
Cx
CCu
duu
dxxx
+=+=
= ∫∫
|ln|lnconstantany is where,||ln
1ln1
36. Let .2+= xu ,1=′u dxdxu =′ and .2−= ux
Cxx
Cuu
C
Cuu
duuu
duuudxxx
++−+=
+−=
+−=
−=
−=+
•••
∫
∫∫
23
25
23
25
23
25
21
23
)2(4)2(56
456
constantany is where
,3223
523
)2(3
)2(323
37. Let .1+= xu ,1=′u dxdxu =′ and .1−= ux
1
1
11
1 where, |1|ln |1|ln1
constant any is where, ||ln
11
11
CCCxxCxx
CCuu
duu
duu
udxx
x
+=++−=++−+=
+−=
⎟⎠⎞
⎜⎝⎛ −=
−=
+
∫
∫∫
38. Let .13 += xu
,3=′u dxdxu 3=′ and .3
1−=
ux
,)13(94)13(
274
constantany is where
,232
92
192
131
312
132
21
23
1
121
23
Cxx
C
Cuu
duu
u
duu
udxxx
++−+=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−=
+
∫
∫∫ ••
where 192 CC +=
39. Let .4+= tu ,1=′u dtdtu =′ and .4−= ut
Ctt
CCuu
duuu
duuudttt
++
−+
=
+−=
−=
−=+
∫∫∫
3)4(
13)4(
constantany is where,313
)4(
)4()4(
1213
1213
1112
1111
40. Let .2 tu −= ,1−=′u dtdtu −=′ and .2 ut −=
Ctt
CCuu
duuu
duu
u
duuudt
tt
+−+−
−=
++−=
−=
−−=
+−=
−+
−−
−−
−−∫
∫
∫∫
32
32
43
4
44
)2(2
)2(
constantany is where,2
)3(
3
12)2(1
© Hong Kong Educational Publishing Co. 156
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
41. Let .1+= xu Then xu =−1
,2
1x
u =′ dxx
dxu2
1=′
1
1
1
1
2 where, |1|ln22
|1|ln222
constantany is where,||ln22
112
12
21
12
11
CCCxx
Cxx
CCuu
duu
duu
u
dxxx
xdxx
+=++−=
++−+=
+−=
⎟⎠⎞
⎜⎝⎛ −=
−=
+=
+
∫
∫
∫∫ •
42. Let .1ln += xu
,1x
u =′ .1 dxx
dxu =′
constantany is where,4
)1(ln22
121
21ln
2
2
CCx
u
ududxx
x
++
=
=
=+
•
∫∫
43. Let .12 += xu ,2xu =′ .2xdxdxu =′
∫ ++ dxxx )11( 2
1
23
22
1
23
22
1123
21 where,
3)1(
2
3)1(
21
constantany is where,32
21
21
)1(21
CCCxx
Cxx
CCuu
duu
+=++
+=
++
++
=
++=
+=
•
∫
44. Let .72 += xu
,2=′u dxdxu 2=′ and .2
7−=
ux
∫ −+ dxxxx )72(
,3
)72(67
10)72(
constantsany are and where
,33
247
52
41
)7(41
27
21
72
3232
5
21
2
3
123
25
221
23
2
2
Cxxx
CC
CxCuu
dxxduuu
dxxduuu
dxxdxxx
+−+−+
=
−−−−=
−−=
−−
=
−+=
••
∫∫
∫ ∫
∫∫
where )( 21 CCC +−=
45. (a) 3)52( +x
125150608
55)2()5()2()2(23
3232
231
3
+++=
+++=
xxx
xCxCx
∫ + dxx 3)52(
constantany is where
,12575202
)125150608(234
23
C
Cxxxx
dxxxx
++++=
+++= ∫
(b) Let .52 += xu ,2=′u .2dxdxu =′
Cx
CCu
duudxx
++
=
+=
=+
•
∫∫
8)52(
constantany is where,42
121)52(
4
4
33
(c) Yes, the difference between them is a constant. They represent the same group of functions.
6.3 pp.240 – 243
p.240
1. ∫ −= dxxy )38(
constantany is where,34
382 CCxx
dxxdx
+−=
−= ∫∫
Q The curve passes through ).1 ,2( −
11
)2(3)2(41 2
−=+−=−
CC
∴ The equation of the curve is .1134 2 −−= xxy
157 © Hong Kong Educational Publishing Co.
Indefinite Integrals
2. ∫ ⎟⎠⎞
⎜⎝⎛ += dx
xy 12
constantany is where,||ln2
12
CCxx
dxdxx
++=
+= ∫∫
Q The curve passes through ).4 ,1(
∴ C++= 11ln24 3=C ∴ The equation is .3 ||ln2 ++= xxy
3. ∫ += dxey x )2(
constantany is where,2
2
CCxe
dxdxex
x
++=
+= ∫∫
Q The curve passes through ).2 ,0(
∴ Ce ++= )0(22 0
1=C ∴ The equation is .12 ++= xey x
4. ∫ += dxxy 2)21(
Let .21 xu += ,2=′u .2dxdxu =′
∴ ∫= duuy 2
21
Cx
CCu
++
=
+= •
6)21(
constantany is where,32
1
2
3
Q The curve passes through ).0 ,1(
∴ C++
=6
)]1(21[02
29
−=C
∴ The equation is .29
6)21( 3−
+=
xy
5. (a) ∫ += dtty )4.02(
constantany is where,4.02 CCtt ++=
When t = 0, x = 6. ∴ C++= )0(4.006 2
6=C ∴ 64.02 ++= ttx
(b) When t = 4,
6.23
6)4(4.042
=++=x
∴ The amount is 23.6 units.
6. (a) ∫ −= dteD t6.02
Ce
CCe
t
t
+−=
+−
=
−
−
6.0
6.0
310
constantany is where,6.0
2
When t = 0, .35
=D
53
1035
310
35 0
=
+=
+−=
C
Ce
∴ 53
10 6.0 +−= − teD
(b) When t = 4,
53
10 4.2 +−= −eD
7.4= (cor. to 1 d. p.) ∴ The concentration is .g/cm 7.4 3
7. ∫= dten t1.0300
constantany is where,3000 1.0 CCe t +=
When t = 0, n = 12 000.
90003000000 12 0
=+=
CCe
∴ 90003000 1.0 += ten
When t = 8, 90003000 )8(1.0 += en 677 15= (cor. to the nearest integer)
∴ The number of probiotics is 15 677. 8. ∫ −= dteP t3.015.0
constantany is where,5.0 3.0 CCe t +−= −
When t = 0, P = 6.0.
5.6
5.00.6 0
=+−=
CCe
∴ 5.65.0 3.0 +−= − teP
When t = 4, 5.65.0 2.1 +−= −eP 3.6= (cor. to 1 d. p.)
∴ The pH value is 6.3.
© Hong Kong Educational Publishing Co. 158
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
9. ∫ += dt
tR
2310
Let .23 += tu ,3=′u .3dtdtu =′
constantany is where,|23|ln
310
13
10
CCt
duu
R
++=
= ∫
When t = 0, R = 0.5.
2ln
3105.0
2ln3
105.0
−=
+=
C
C
∴ 2ln3
105.0 |23|ln3
10−++= tR
When t = 4,
2ln3
105.0|2)4(3|ln3
10−++=R
0.7= (cor. to 1 d. p.)
∴ The radius of the polluted area is 7.0 m. 10. ∫ −= dteN t4.048
constantany is where,120 4.0 CCe t +−= −
When t = 0, N = 300.
420
120300 0
=+−=
CCe
∴ 420120 4.0 +−= − teN
When t = 4, 420120 6.1 +−= −eN 396= (cor. to the nearest integer)
∴ There are 396 visitors after 4 hours. 11. ∫ −= dteN t2.08
constantany is where,40 2.0 CCe t +−= −
When t = 0, N = 10.
50
4010 0
=+−=
CCe
∴ 5040 2.0 +−= − teN
When t = 7, 5040 4.1 +−= −eN 40= (cor. to the nearest integer)
∴ His typing speed is 40 words per minute.
12. ∫−= dttM 53
8
constantany is where,5 58
CCt +−=
When t = 0, M = 6000.
6000
)0(56000=
+−=C
C
∴ 60005 58
+−= tM
When M = 1000,
9894.741000
1000
600051000
85
58
58
≈=
=
+−=
t
t
t
∴ It will take 75 months for the price to decrease by half.
13. ∫ += dttP )400300(
constantany is where,3
800300 23
CCtt ++=
When t = 0, P = 5000.
5000
605000=
++=C
C
∴ 50003
800300 23
++= ttP
When t = 2,
6354
500023
8002300 23
=
+×+×=P
∴ The population after 2 years is 6354.
p.242
14. 12
2+= xe
dxyd
constantany is where,
)1(
11 CCxe
dxedxdy
x
x
++=
+= ∫
When x = 0, 1−=dxdy .
2
01
1
10
−=++=−
CCe
∴ 2−+= xedxdy x
constantany is where,2
2
)2(
22
2CCxxe
dxxey
x
x
+−+=
−+= ∫
Since the point (0, 6) is on the curve,
5
006
2
20
=+−+=
CCe
∴ The equation of the curve is
.522
2+−+= xxey x
159 © Hong Kong Educational Publishing Co.
Indefinite Integrals
15. xdx
yd=2
2
constantany is where,2 11
2CCxxdx
dxdy
+== ∫
constantany is where,6
2
221
3
1
2
CCxCx
dxCxy
++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+= ∫
Q The curve passes through (0, 3) and (2, 9).
37
32629
3003
1
1
32
2
=
++=
=++=
C
C
CC
∴ The equation of the curve is
.337
6
3++= xxy
16. 24 22
2+= xe
dxyd
constantany is where,22
)24(
112
2
CCxe
dxedxdy
x
x
++=
+= ∫
Q The slope of the tangent at (0, 1) is 4.
40=
=xdxdy
∴ 10 024 Ce ++=
21 =C
∴ 222 2 ++= xedxdy x
constantany is where,2
)222(
2222
2
CCxxe
dxxeyx
x
+++=
++= ∫
Since the curve passes through (0, 1), ∴ 2
0 001 Ce +++= 02 =C
∴ The equation of the curve is .222 xxey x ++=
17. xdx
yd−= 32
2
constantany is where,
23
)3(
11
2CCxx
dxxdxdy
+−=
−= ∫
Since the slope of the tangent y = x + 1 is 1,
11=
=xdxdy
23
21)1(31
1
1
−=
+−=
C
C
∴ 23
23
2−−=
xxdxdy
constantany is where,23
623
23
23
12
32
2
CCxxx
dxxxy
+−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−= ∫
Since (1, 2) is on the curve,
∴ 223
61
232 C+−−=
6
132 =C
∴ The equation of the curve is
.6
1323
23
61 23 +−+−= xxxy
18. ∫ += dt
ttn
1
Let .1+= tu Then .1−= ut 1=′u and dtdtu =′
Ctt
CCuu
duuu
duu
un
++−+=
+−=
−=
−=
∫
∫−
21
23
21
23
21
21
)1(2)1(32
constantany is where,232
)(
1
When t = 0, n = 36.
3137
23236
=
+−=
C
C
∴ 31
21
23
37)1(2)1(32
++−+= ttn
When t = 5,
3137)15(2)15(
32 2
123
++−+=n
42= (cor. to the nearest integer) Q The number of leopards after 5 years is 42. 19. (a) dtel t∫ −= 3.01.2
constantany is where,7 3.0 CCe t +−= −
When t = 0, l = 1.
8
71 0
=+−=
CCe
∴ 87 3.0 +−= − tel
© Hong Kong Educational Publishing Co. 160
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
(b) When t = 10, 87 3 +−= −el 7.7= (cor. to 1 d. p.)
∴ The length of the prawn after ten days is 7.7 cm. (c) When l = 6,
1759.43.0
72ln
72
876
3.0
3.0
≈−
=
=
+−=
−
−
t
t
e
e
t
t
∴ 5 days are needed before they can be sold.
20. (a) dtttP ∫ += 42 2
Let .42 2 += tu ,4tu =′ .4tdtdtu =′
Ct
CCu
duuP
++=
+=
=
•
∫
23
2
23
)42(61
constantany is where,32
41
41
(b) When t = 4, P = 102.
66
]4)4(2[61102 2
32
=
++=
C
C
∴ 66)42(61 2
32 ++= tP
When t = 5,
66]4)5(2[61 2
32 ++=P
132= (cor. to the nearest integer)
∴ The profit after 5 years is 132 thousand dollars. 21. (a) dtteP t∫ −−=
2
10000
Let .2tu = ,2tu =′ .2tdtdtu =′
Ce
CCe
dueP
t
u
u
+=
+−−=
−=
−
−
−• ∫
2
5000
constantany is where,)(50002110000
When t = 0, P = 6000. ∴ Ce += 050006000 1000=C
∴ 100050002
+= −teP
(b) When t = 5, 10005000
25 += −eP 1000= (cor. to the nearest dollars)
∴ The value is $1000.
22. (a) dtttt
P ∫ +++
= 92
50 2
dtttdtt ∫∫ +++
= 92
150 2
For the second integral, Let .92 += tu ,2tu =′ .2tdtdtu =′
duudttt ∫∫ =+ 21
2
219
∴ duudtt
P ∫∫ ++
= 21
21
2150
Ctt
C
Ctt
++++=
++++= •
23
2
23
2
)9(31|2|ln50
constantany is where
,)9(32
21|2|ln50
When t = 0, P = 10.5.
2ln505.1
)9(312ln505.10 2
3
−=
++=
C
C
∴ 2ln505.1)9(31|2|ln50 2
32 −++++= ttP
(b) The number of year from 2009 to 2016
7
20082016=
−=
2ln505.1)97(31)27ln(50 2
32 −++++=P
9.223= (cor. to 1 d. p.)
∴ The annual profit in 2016 is 2239 thousand dollars.
161 © Hong Kong Educational Publishing Co.
Indefinite Integrals
pp.248 – 251
p.248
1. ∫∫∫ −− +=+ dxxdxxdxxx 2323 34)34(
Cxx
C
Cxx
+−=
+−+⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
14
14
3
constantany is where
,)(34
4
2. ∫∫∫−
−=⎟⎟⎠
⎞⎜⎜⎝
⎛− dxxdxxdx
xx 2
121
212
Cxx
C
Cxx
+−=
+−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
21
23
21
23
234
constantany is where
,)2(322
3. ∫∫∫ +=+ −− dxdxedxe xx 63)63( 22
constantany is where
,62
3 2
C
Cxe x++−=
−
4. ∫∫ −=−+ dttdttt )1()1)(1( 2
constantany is where,3
3CCtt +−=
5. dtttdttt )2()21( 322 ∫∫ +=+
Ctt
CCtt
++=
+⎟⎟⎠
⎞⎜⎜⎝
⎛+=
23
constantany is where,4
23
43
43
6. ∫∫++
=+ dt
tttdt
tt
3
2
3
2 12)1(
Cttt
C
Cttt
dtttt
+−−=
+⎟⎟⎠
⎞⎜⎜⎝
⎛
−+−+=
++=
−−
−−
−−−∫
22 ||ln
constantany is where
,2
)(2 ||ln
)2(
21
21
321
7. Let .23 −= xu ,3 2xu =′ .3 2dxxdxu =′
∫ ∫=− duudxxx 5532 )2(3
Cx
CCu
+−
=
+=
6)2(
constantany is where,6
63
6
8. Let .14 += xu ,4 3xu =′ .4 3dxxdxu =′
∫ ∫=+ duedxex ux 1 3 4
4
Ce
CCex
u
+=
+=+14
constantany is where,
9. Let .1 2xu −= ,2xu −=′ .2xdxdxu −=′
∫ ∫−=− dxudxxx 442
21)1(
Cx
CCu
+−
−=
+−= •
10)1(
constantany is where,52
1
52
5
10. Let .1 xeu −= ,xeu −=′ .dxedxu x−=′
∫ ∫−=− dxudxee xx 1
Ce
CCu
x +−−=
+−=
23
23
)1(32
constantany is where,32
11. Let .xx eeu −+= ,xx eeu −−=′ .)2( dxedxu xx −−=′
∫ ∫=+−
−
−
duu
dxeeee
xx
xx 1
Cee
CCuxx ++=
+=− ||ln
constantany is where,||ln
12. ∫∫ +++
=+++ dx
xxxdx
xxx
1)1(
11
2
2
2
2
∫ ∫ ++= dx
xxdx
12
For the second integral, let ,12 += xu ,2xu =′ .2xdxdxu −=′
∫ ∫=+
duu
dxx
x 121
12
∴ ∫ ∫∫ +=+++ du
udxdx
xxx 1
21
11
2
2
Cxx
C
Cux
+++=
++=
|1|ln21
constantany is where
,||ln21
2
© Hong Kong Educational Publishing Co. 162
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
13. ∫∫ −+=+
= dxxdxx
xy )1(1 22
2
constantany is where,1 CCx
x +−=
Q The curve passes through (1, 2).
∴ C+−=1112
2=C
∴ 21+−=
xxy
14. ∫ += dx
xy
141
Let .14 += xu ,4=′u .4dxdxu −=′
∴ ∫= duu
y 141
Cx
CCu
++=
+=
|14|ln41
constantany is where,||ln41
When x = 2, y = 5.
9ln
415
9ln415
−=
+=
C
C
∴ 9ln415|14|ln
41
−++= xy
59
14ln41
++
=x
15. (a) ∫ +−= dttd )41.0(
Ctt
CCtt
++−=
+⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
2
2
21.0
constantany is where,2
41.0
When t = 0, d = 0. ∴ 0=C ∴ 221.0 ttd +−=
(b) When t = 10,
199
2001)10(2)10(1.0 2
=+−=
+−=d
∴ The distance travelled is 199 m.
p.248
16. Let .2 xu −= Then .2 ux −= ,1−=′u .dxdxu −=′
∫ ∫ +−−=−+ duuudxxx )12(2)1(
Cxx
C
Cuu
duuu
duuu
+−−−=
+−=
−=
−−=
•
∫
∫
23
25
23
25
21
23
)2(2)2(52
constantany is where
,323
52
)3(
)3(
17. Let .1+= tu Then .1−= ut ,1=′u .dtdtu −=′
∫ − dttt 13
Ctttt
C
Cuuuu
duuuuu
duuuuu
duuu
++−+++−+=
+−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−=
−+−=
−+−=
−−=
∫
∫∫
−
23
25
27
29
23
25
27
29
21
23
25
27
123
3
)1(32)1(
56)1(
76)1(
92
constantany is where
,32
523
723
92
)33(
)133(
)1(
18. Let .3+= xu Then .3−= ux ,1=′u .dxdxu =′
∫ ∫ −−=+ duuudxxx 100100 )3()3(
Cxx
CCuu
duuu
++
−+
=
+−=
−= ∫
101)3(3
102)3(
constantany is where,101
3102
)3(
101102
101102
100101
19.
dxe
e
dxee
edx
e
x
x
x
x
xx
∫
∫∫
+=
+=
+•
−−
1
11
11
2
2
2
2
22
Let .12 += xeu ,2 2xeu =′ xedxu 22=′ .
∫ ∫=+ −
duu
dxe x 1
21
11
2
Ce
C
Cu
x ++=
+=
|1|ln21
constantany is where
,||ln21
2
163 © Hong Kong Educational Publishing Co.
Indefinite Integrals
20. Let .12 += xu Then .12 −= ux ,2xu =′ .2xdxdxu =′
∫ ∫+
=+
•dx
x
xx
x
x 12
1
12 2
2
2
3
121)1(
61
constantany is where
,232
41
)(41
121
21
223
2
21
23
21
21
Cxx
C
Cuu
duuu
duu
u
++−+=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−=
−=
−=
∫
∫−
•
21. Let .5ln xu =
,1x
u =′ .x
dxdxu =′
duudxx
x 5ln∫ ∫=
Cx
CCu
+=
+=
2
2
)5(ln21
constantany is where,2
22. dxx
xx∫ ⎥⎦
⎤⎢⎣
⎡
+++
)4(3)12( 2
4
∫ ∫ +++= dx
xxdxx
43)12( 2
4
For the first integral, let .12 += xu ,2=′u .2dxdxu =′
For the second integral, let .42 += xu ,2xu =′ .2xdxdxu =′
∴ dxx
xx∫ ⎥⎦
⎤⎢⎣
⎡
+++
)4(3)12( 2
4
Cxx
CCuu
duu
duu
++++
=
++=
+=
•
∫∫
|4|ln23
10)12(
constantany is where,||ln23
521
123
21
25
5
4
23. ∫ ∫∫ −=⎥⎦⎤
⎢⎣⎡ − −− dx
xxdxxedx
xxxe xx lnln 22
For the first integral, let .2xu −= ,2xu −=′ .2xdxdxu −=′
For the second integral, let .ln xu =
,1x
u =′ .1 dxx
dxu =′
∴ dxxxxe x∫ ⎥⎦⎤
⎢⎣⎡ −− ln2
Cxe
CCue
ududue
x
u
u
+−−=
+−−=
−−=
−
∫∫
2|)|(ln
21
constantany is where,22
121
2
2
2
24. dttth ∫ ++
=1
)1ln(
Let ).1ln( += tu
,1
1+
=′t
u .1
1 dtt
dtu+
=′
∴ duuh ∫=
Ct
CCu
++
=
+=
2)]1[ln(
constantany is where,2
2
2
When t = 0, h = 2
22
)10ln(22
=
+⎟⎠⎞
⎜⎝⎛ +
=
C
C
∴ 22
)]1[ln( 2+
+=
th
25. dxe
exf x
x
∫+
= 21)(
constantany is where,
21
)(
2
2
CCee
dxee
xx
xx
+−−=
+=
−−
−−∫
211
421
4)0(
00
=
=+−−
=
C
Cee
f
∴ 2
1121)( 2 +−−= −− xx eexf
© Hong Kong Educational Publishing Co. 164
New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6
26. dxx
exfx
∫=)(
Let .xu =
,2
1x
u =′ .2
1 dxx
dxu =′
∴ ∫= duexf u2)(
Ce
CCex
u
+=
+=
2
constantany is where,2
2
24
2
2
)4(
eC
eCe
ef
−=
=+
=
∴ 22)( eexf x −=
27. (a) 2)2(
2 +++
=+
+x
BxAx
BA
2
2+
++=
xBAAx
Q 2
22 +
++≡
+ xBAAx
xx
∴ 1=A
∴ 2
0)1(202
−==+=+
BBBA
(b) ∫ ∫ ⎟⎠⎞
⎜⎝⎛
+−=
+dx
xdx
xx
221
2
Cxx ++−= |2|ln2 ,
where C is any constant
28. (a) )1(
)1(1 +
++=
++
xxBxxA
xB
xA
)1(
)(+++
=xx
AxBA
Q )1(
)()1(
1+++
≡+ xx
AxBAxx
∴ 1=A
10−=
=+BBA
(b) ∫ +dx
xx )1(1
constantany is where,|1|ln||ln
111
CCxx
dxxx
++−=
⎟⎠⎞
⎜⎝⎛
+−= ∫
29. (a) )12)(1(
)1()12(121 −+
++−=
−+
+ xxxBxA
xB
xA
)12)(1()2(
)12)(1(2
−+−++
=
−+++−
=
xxABxBA
xxBBxAAx
Q )12)(1(
)2()12)(1(
3−+−++
≡−+ xx
ABxBAxx
x
∴ ⎩⎨⎧
=−=+
)2....(..........0)1...(..........32
ABBA
:)2()1( −
133
==
AA
Substituting A = 1 into (2),
101
==−
BB
(b) ∫ −+ )12)(1(3
xxx
∫∫
∫
−+
+=
⎟⎠⎞
⎜⎝⎛
−+
+=
dxx
dxx
dxxx
121
11
121
11
For the second integral, Let .12 −= xu ,2=′u .2dxdxu =′
∴ ∫ −+ )12)(1(3
xxx
Cxx
C
Cux
duu
dxx
+−++=
+++=
++
= ∫ ∫
|12|ln21|1|ln
constantany is where
,||ln21|1|ln
121
11
30. (a) 32 −
++ x
Bx
A
)3)(2(32)(
)3)(2(23
−+−++
=
−+++−
=
xxABxBA
xxBBxAAx
Q )3)(2(32)(
)3)(2(5
−+−++
≡−+ xx
ABxBAxx
∴ ⎩⎨⎧
=−=+
)2....(..........532)1........(..........0
ABBA
:)2(2)1( −×
1555)32(22
−=−=−=−−+
AA
ABBA
Substituting 1−=A into (1),
1010
==+−=+
BBBA
165 © Hong Kong Educational Publishing Co.
Indefinite Integrals
(b) ∫ −+dx
xx )3)(2(5
constantany is where
,|3|ln|2|ln3
12
1
C
Cxx
dxxx
+−++−=
⎟⎠⎞
⎜⎝⎛
−−
+−
= ∫
31. ∫ += dt
ttT 231
Let .31 2tu += ,6tu =′ .6tdtdtu =′
∴ ∫= duu
T 161
Ct
CCu
++=
+=
|31|ln61
constantany is where,||ln61
2
When t = 0, T = 25.
25
|ln|6125
=
+=
C
C
∴ 25|31|ln61 2 ++= tT
When t = 10,
25|)10(31|ln61 2 ++=T
0.26= (cor. to 1 d. p.)
∴ The average temperature after ten years is C.0.26 °
32 – 33 (GCE Questions) 34 – 35 (HKCEE Questions) 36 – 38 (HKASLE Questions) Extended Questions p.251
39. Let ,)( 2 cbxaxxf ++=′
where a, b and c are constants. From the graph, .1)0( =′f ∴ cba ++= )0()0(1 2
∴ 1=c ∴ 1)( 2 ++=′ bxaxxf From the graph, 0)1( =′f and .0)1( =−′f
⎩⎨⎧
=+−=++
)2....(..........01)1.....(..........01
baba
:)2()1( −
002
==
bb
Substituting b = 0 into (1),
1
010−=
=++a
a
∴ 1)( 2 +−=′ xxf
constantany is where,
3
)1()(3
2
CCxx
dxxxf
++−=
+−= ∫
Q 1)0( =f
∴ C++= 03
013
∴ 1=C
∴ 13
)(3
++= xxxf
Open-ended Question p.251
40. Consider .)2()( 10+= xxf
.)2(10)]([ 9+= xxfdxd
∴ ∫ ++=+ 1109 )2()2(10 Cxdxx ,
where 1C is any constant
∫∫
∫
++++
=
++
=+++
dxxdxxx
CC
Cxdxxx
88
12
2
108
)2()2)(1(10
where
,10
)2()11()2(
2
10
10)2( Cx
++
=
∫ ++
+++ 3
98
9)2()2)(1( Cxdxxx ,
where 3C is any constant
2
10
10)2( Cx
++
=
∴ ∫ ++
−+
=++ Cxxdxxx9
)2(10
)2()2)(1(910
8
where 32 CCC −=
(or other reasonable answers)