BAB I
9. β«π₯+1
π₯2 β4π₯+8 =
1
2β«
2(π₯+1)
π₯2β4π₯+8ππ₯ =
1
2 β«
2π₯+2
π₯2β4π₯ +8ππ₯ =
1
2β«
(2π₯β4)+6
π₯2 β4π₯+8ππ₯
= 1
2β«
2πβ4
π₯2β4π +8ππ₯ +
1
2 β«
6
π₯2 β4π₯+8ππ₯
= 1
2 ln (π₯ 2 β 4π₯ + 8) +
1
2 . 6β«
ππ₯
π₯2β4π₯+8
= 1
2 ln (π₯ 2 β 4π₯ + 8) +3β«
ππ₯
(π₯β2)2+(β4)2
= 1
2 ln (π₯ 2 β 4π₯ + 8) + 3.
1
2πππ tan
π₯β2
2+ π
13. β«(5β4π₯)ππ₯
β12π₯ β4π₯2β8 =
1
2β«
2 (5β4π₯) ππ₯
β12π₯ β4π₯2β8ππ₯ =
1
2β«
10 β8π₯
β12π₯ β4π₯2 β8ππ₯
= 1
2β«
β2π₯+(12β8π₯ )
β12π₯ β4π₯2β8ππ₯
= 1
2β«
β2
β12π₯ β4π₯2β8ππ₯ +
1
2β«
12β8π₯
β12π₯ β4π₯2β8ππ₯
= 1
2β«
β2ππ₯
β(1)2β(2π₯ β3)2+ β12π₯ β 4π₯2 β 8 + π
= 1
2. β1 β«
2ππ₯
β(1)2β(2π₯ β3)2+ β12π₯ β 4π₯2 β 8 + π
= β1
2 πππ sin
2π₯ β3
1+ β12π₯ β 4π₯ 2 β 8 + π
15. β«(π₯β1)ππ₯
3π₯2β4π₯ +3 =
1
6β«
6(π₯β1)ππ₯
3π₯2β4π₯ +3=
1
6β«
6π₯β6
3π₯2 β4π₯+3ππ₯ =
1
6β«
(6π₯β4)β2
3π₯2β4π₯ +3ππ₯
= 1
6β«
6π₯ β4
3π₯2β4π₯ +3ππ₯ +
1
6β«
β2
3π₯2β4π₯ +3ππ₯ =
1
6ln(3π₯2 β 4π₯ +
3) +1
6. β2 β«
ππ₯
3π₯2β4π₯+3
= 1
6ln(3π₯2 β 4π₯ + 3) β
1
3β«
ππ₯
(β5)2
β(3π₯β2)2
= 1
6ln(3π₯2 β 4π₯ + 3) β
1
3.
1
β5 πππ tan
3π₯ β2
β5+ π
BAB II
7. β« πππ tan π₯ ππ₯ = π₯ πππtan π₯ β ln β1 + π₯2 + π
β« πππ tan π₯ ππ₯ = uvβ β« π£ππ’
U= arc tan x dv = dx du = 1
1+π₯2 ππ₯ v = x
β« πππ tan π₯ ππ₯ = πππ tan π₯. π₯ β β« π₯.1
1+π₯2 ππ₯
= π₯ πππ tan π₯ β 1
2β«
2π₯
1+π₯2 ππ₯
= π₯ πππ tan π₯ β1
2ln|1 + π₯2| + π
= π₯ πππ tan π₯ β ππ β1 + π₯2 + π
9.β« πππ π π’ππ’ =πππ πβ1π’ π πππ’
π+
πβ1
πβ« πππ πβ2π’ππ’
β« πππ ππ’ππ’ = β« πππ πβ1+1π’ππ’ = β« πππ πβ1 π’ πππ π’ππ’
π¦ = πππ πβ1π’
ππ¦ = π β 1πππ πβ2π’ β sin π’ = β(π β 1).πππ πβ2 sin π’ ππ’
dx= cos u du
x= sin u
β β« πππ ππ’ππ’ = πππ πβ1 π’ sin π’ β β« sin π’. βπ β 1. πππ πβ1
= πππ πβ1 sin π’ + π β 1 β« π ππ2 π’. πππ πβ2ππ’
= πππ πβ1 sin π’ + π β 1 β«(1 β πππ 2π’).πππ πβ2 π’ ππ’
= πππ πβ1 sin π’ + π β 1 β« πππ πβ2 β πππ ππ’ ππ’
= πππ πβ1 sin π’ + π β 1 β« πππ πβ2 β (π β 1) β« πππ π π’ ππ’
= πππ πβ1 sin π’ + π β 1 β« πππ πβ2 β π β« πππ ππ’ ππ’ + β« πππ ππ’ ππ’
β« πππ ππ’ ππ’ + π β« πππ ππ’ ππ’ β β« πππ π π’ ππ’ = πππ πβ1 sin π’ + π β
1 β« πππ πβ2ππ’
mβ« πππ π π’ ππ’ = πππ πβ1 sin π’ + π β 1 β« πππ πβ2ππ’
β« πππ ππ’ ππ’ =πππ πβ1 sin π’
π+
πβ1
πβ« πππ πβ2π’ ππ’
REDUKSI
1. β«ππ₯
(1βπ₯2)3 = 1
12 {π₯
(2.3β2)(12βπ₯2)3β1 +2.3β3
2.3β2β«
ππ₯
(1βπ₯2)3β2
=π₯
4(12βπ₯2)2 +3
4β«
ππ₯
(1βπ₯2 )2
=π₯
4(12βπ₯2)2 +3
4{
π₯
(2.2β2) (1βπ₯2)+
2.2β3
2.2β2β«
ππ₯
(1βπ₯2)
=π₯
4(12βπ₯2)2 +3
4{
π₯
2(1βπ₯2)+
1
2.
1
2ln |
1+π₯
1βπ₯| + π
=π₯
4(12βπ₯2)2 +3π₯
8(1βπ₯2)+
3
16ln |
1+π₯
1βπ₯| + π
=2π₯
8(1βπ₯2)2 +3π₯(1βπ₯2)
8(1βπ₯2)2 +3
16ln |
1+π₯
1βπ₯| = π
=5π₯ β3π₯3
8(1βπ₯2)2 +3
16ln |
1+π₯
1βπ₯| + π
=π₯(5β3π₯2)
8(1βπ₯2)2 +3
16ln |
1+π₯
1βπ₯| + π
3. β« πππ 5π₯ ππ₯ = 1
5(3πππ 4π₯ + 4πππ 2π₯ + 8)π πππ₯ + π
Rumus reduksi 7 β πππ ππ₯ ππ₯ =πππ πβ1π₯π πππ₯
π+
πβ1
πβ« πππ πβ2 π₯ ππ₯
β« πππ 5π₯ ππ₯ = πππ 5 β1π₯π πππ₯
5+
5β1
5β« πππ 5β2 π₯ ππ₯
β« π¦ππ₯ = π¦π₯ β β« π₯ππ¦
=πππ 5β1π₯π πππ₯
5+
4
5β« πππ 3π₯ ππ₯
=πππ 4 π πππ₯
5+
4
5{
πππ 2 π₯π πππ₯
3+
2
3β« πππ π₯ ππ₯}
=πππ 4 π₯π πππ₯
5+
4πππ 2 π₯π πππ₯
15
8
15π πππ₯
=1
15{3πππ 4π₯π πππ₯ + 4πππ 2π₯π πππ₯ + 8π πππ₯} + π
=1
15{3πππ 4π₯ + 4πππ 2π₯ + 8}π πππ₯ + π
5. β« π2π₯ (2 sin 4π₯ β 5 πππ 4π₯)ππ₯ =1
25π2π₯(β14 sin 4π₯ β 23 cos 4π₯) + π
π2π₯ (π΄ cos 4π₯ + π΅ sin 4π₯) β ππππ:
π3π₯ (2 sin 4π₯ β 5 cos 4π₯) = π3π₯ (3π΄ + 4π΅)πΆππ 4π + π3π₯ (3π΅ β
4π΄) sin 4π₯
Maka:
3A+4B=-5 x3 9A+12B=-15 -4A+3B=2 x4 -16A=12B=8 _
25A= -23 A= -23/25
3π΅ = 2 + 4π΄ = 2 β92
25=
β42
25β π΅ =
β14
25
β πΈ2π(2 sin 4π₯ β 5 cos 4π₯)ππ₯
= π2π₯ (β23
25πππ 4π₯ β
14
25sin 4π₯) + π
= β1
25π2π₯ (23 cos 4π₯ + 14 sin 4π₯) + π
=1
25π2π₯ (β14 sin 4π₯ β 23 πππ 4π₯) + π
BAB III
10. β« (sec π₯
tan π₯)
4
ππ₯ =1
3 π‘ππ3 π₯β
1
tan π₯+ π
= β« (1
πππ π₯π πππ₯
πππ π₯
)
4
ππ₯ = β« (1
πππ π₯.
πππ π₯
π πππ₯)
4
ππ₯ = β« (1
π πππ₯)
4
ππ₯
= β« πππ ππ4π₯ππ₯ = β« πππ ππ2 (1 + πππ‘2 π₯)ππ₯
= β« πππ ππ2ππ₯ + β« πππ ππ2π₯πππ‘2 π₯ ππ₯
= βπππ‘π₯ + β« πππ‘2π₯πππ ππ2 π₯ ππ₯
= βπππ‘π₯ + β« πππ‘2π₯ π(βπππ‘π₯)
= βπππ‘π₯ + β1
3πππ‘3π₯ + π
= β1
3π‘ππ3 π₯β
1
π‘πππ₯+ π
13.
15. β«πππ‘3 π₯ ππ₯
πππ ππ π₯ = β«
πππ‘2 π₯.πππ‘π₯ ππ₯
πππ ππ π₯ = β«
(πππ ππ2 π₯β1).cot π₯ ππ₯
πππ ππ π₯
= β«πππ ππ2 π₯.cot π₯
πππ ππ π₯ππ₯ β β«
cotπ₯
πππ ππ π₯ππ₯
= β« πππ ππ π₯. cotπ₯ ππ₯ β β« cos ππ₯
= -cosec x β sin x+c = -sin x- cosec x +c
BAB IV
9. β«β25 βπ₯2
π= 5 ln |
5ββ25 βπ₯2
π₯| + β25 β π₯2 + π
a=5 b=1 u= x π’ =π
πsin π§ π₯ =
5 π πππ§
dx= 5 cos z sin π§ =π₯
5
β25 β π₯2 = π cos π§ = 5 cos π§
β«β25 βπ₯2
πππ₯ = β«
(5 cos π§)(5 cos π§)
5 sin π§ = β«
25 πππ 2π§
5 sin π§ππ§
= 5 β«πππ 2 π§
sin π§ππ§ = 5 β«
(1βπ ππ2 π§)
sin π§ππ§
= 5(β«1
π πππ§ππ§ β β« sin π§ ππ§
= 5 β« πππ ππ π§ ππ§ β 5 β« sin π§ ππ§
= 5|πππ ππ π§ β cotπ§| + 5 cos π§ + π
= 5|5
π₯β
β25βπ₯2
π₯| + 5.
β25βπ₯2
π₯+ π
= 5 |5ββ25βπ₯2
π₯| + β25 β π₯2 + π
13.
BAB V
12. β«π₯2+3π₯β4
π₯2β2π₯β8ππ₯ = π₯ + ln|(π₯ + 2)(π₯ β 4)4| + π
β π₯ 2 β 2π₯ β 81
βπ₯2+3π₯ β4π₯2β2π₯ β8
5π₯ +4β
β β« 1 +5π₯ +4
π₯2β2π₯β8ππ₯
βπ₯ 2 β 2π₯ β 8 = (π₯ + 2)(π₯ β 4)
5π₯ +4
π₯2 β2π₯β8=
π΄
π₯+2+
π΅
π₯β4 dikali (x+2)(x-4)
5x+4 = A(x-4)+B(x+2) = (A+B)x-4A+2B
A=1,B=4
ββ« 1 +5π₯ +4
π₯2 β2π₯β8ππ₯ = β« 1ππ₯ + β«
π΄ ππ₯
(π₯+2)+ β«
π΅ ππ₯
(π₯β4)
= π₯ + β«ππ₯
(π₯+2)+ 4 β«
ππ₯
π₯β4
= π₯ + ln(π₯ + 2) + 4 ln(π₯ β 4) + π
= π₯ + ln|(π₯ + 2)(π₯ β 4)4| + π