Tugas Matematika

1. Tugas Matematika
Integral Hal 49- 59
Disusun Oleh :
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TAHUN AJARAN 2014/2015
Industri Air Kantung Sungailiat 33211
Bangka Induk, Propinsi Kepulauan Bangka Belitung
Telp : +62717 93586
Fax : +6271793585 email : polman@polman-babel.ac.id
http://www.polman-babel.ac.id
Kelompok 7 :
- Fery Ardiansyah
- Rakam Tiano
- Sarman
- Mirza ramadhan

2. Dua aturan integrasi berguna
Latihan 7.7
Cari integral tak tentu yang paling umum..
1. 3𝑥4
− 5𝑥3
− 21𝑥2
+ 36𝑥 − 10 𝑑𝑥
2. 3𝑥2
− 4𝑐𝑜𝑠 2𝑥 𝑑𝑥
3.
8
𝑡5
+
5
𝑡
𝑑𝑡
4.
1
25 − 𝜃2
+
1
100 + 𝜃2
𝑑𝜃
5.
𝑒5𝑥
− 𝑒4𝑥
𝑒2𝑥
𝑑𝑥
6.
𝑥7
+ 𝑥4
𝑥5
𝑑𝑥
7.
𝑥7
+ 𝑥4
𝑥5
𝑑𝑥
8. 𝑥2
+ 4 2
𝑑𝑥 = 𝑥4
9.
7
𝑡
3 𝑑𝑡
10.
20 + 𝑥
𝑥
𝑑𝑥

3. Penyelesaian :
1. 3𝑥4
− 5𝑥3
− 21𝑥2
+ 36𝑥 − 10 𝑑𝑥 = 3𝑥4
𝑑𝑥 − 5𝑥3
𝑑𝑥 − 21𝑥2
𝑑𝑥 +
36𝑥 𝑑𝑥 − 10 𝑑𝑥 = 3 𝑥4
𝑑𝑥 − 5 𝑥3
𝑑𝑥 − 21 𝑥2
𝑑𝑥 + 36 𝑥 𝑑𝑥 −
10 𝑑𝑥 = 3
𝑥5
5
− 5
𝑥4
4
− 21
𝑥3
3
+ 36
𝑥2
2
− 10𝑥 + 𝑐 =
3
5
𝑥5
−
5
4
𝑥4
− 7𝑥3
+
18𝑥2
− 10𝑥 + 𝑐
2. 3𝑥2
− 4𝑐𝑜𝑠 2𝑥 𝑑𝑥 = 3𝑥2
𝑑𝑥 − 4 𝑐𝑜𝑠 2𝑥 𝑑𝑥 = 3 𝑥2
𝑑𝑥 − 4 𝑐𝑜𝑠 2𝑥 𝑑𝑥 =
3
𝑥3
3
− 4
1
2
𝑠𝑖𝑛2𝑥 + 𝑐 = 𝑥3
− 2 sin 2𝑥 + 𝑐
3.
8
𝑡5 +
5
𝑡
𝑑𝑡 =
8
𝑡5 𝑑𝑥 +
5
𝑡
𝑑𝑥 = 8 𝑡−5
𝑑𝑥 + 5
1
𝑡
𝑑𝑥 = 8
𝑡−4
−4
+ 5 𝑙𝑛 𝑡 + 𝑐 =
−2𝑡−4
+ 5 𝑙𝑛 𝑡 + 𝑐
4.
1
25−𝜃2
+
1
100+𝜃2 𝑑𝜃 =
1
25−𝜃2
𝑑𝑥 +
1
100+𝜃2 𝑑𝑥 =
1
52+𝜃2
𝑑𝑥 +
1
102+𝜃2 𝑑𝑥 =
𝑠𝑖𝑛−1 𝜃
5
+
1
10
𝑡𝑎𝑛−1 𝜃
10
+ 𝑐
5.
𝑒5𝑥 −𝑒4𝑥
𝑒2𝑥 𝑑𝑥 = 𝑒3𝑥
− 𝑒2𝑥
𝑑𝑥 = 𝑒3𝑥
𝑑𝑥 − 𝑒2𝑥
𝑑𝑥 =
1
3
𝑒3𝑥
−
1
2
𝑒2𝑥
+ 𝑐
6.
𝑥7+𝑥4
𝑥5 𝑑𝑥 =
𝑥7
𝑥5 𝑑𝑥 +
𝑥4
𝑥5 𝑑𝑥 =
7.
1
𝑒6+𝑥2 𝑑𝑥 = 𝑒6
+ 𝑥2
𝑑𝑥 = 𝑙𝑛 𝑒6
+ 𝑥2
+ 𝑐
8. 𝑥2
+ 4 2
𝑑𝑥 = 𝑥4
+ 16 + 2. 𝑥2
. 4 𝑑𝑥 = 𝑥4
+ 8𝑥2
+ 16 𝑑𝑥 =
1
4+1
𝑥4+1
+
8
2+1
𝑥2+1
+ 16𝑥 + 𝑐 =
1
5
𝑥5
+
8
3
𝑥3
+ 𝑐
9.
7
𝑡3 𝑑𝑡 = 7𝑡−
1
3 𝑑𝑡 =
7
−
1
3
+1
𝑡−
1
3
+1
+ 𝑐 =
7
2
3
𝑡
2
3 + 𝑐 =
21
2
𝑡
2
3 + 𝑐
10.
20+𝑥
𝑥
𝑑𝑥 = 20 + 𝑥 𝑥−
1
2 𝑑𝑥 = 20𝑥−
1
2 + 𝑥
1
2 𝑑𝑥 =
20
−
1
2
+1
𝑥−
1
2
+1
+
1
1
2
+1
𝑥
1
2
+1
+ 𝑐 =
20
1
2
𝑥
1
2 +
1
3
2
𝑥
3
2 + 𝑐 = 40𝑥
1
2 +
2
3
𝑥
3
2 + 𝑐

4. Integrasi dasar teknik
Integrasi dengan substitusi
Latihan 8.1
Gunakan integrasi dengan substitusi untuk menemukan integral tak tentu yang paling umum.
1. 3 𝑥3
− 5 4
𝑥2
𝑑𝑥
2. 𝑒 𝑥4
𝑥3
𝑑𝑥
3.
𝑡
𝑡2 + 7
𝑑𝑡
4. 𝑥5
− 3𝑥
1
4 5𝑥4
− 3 𝑑𝑥
5.
𝑥3
− 2𝑥
𝑥4 − 4𝑥2 + 5 4
𝑑𝑥
6.
𝑥3
− 2𝑥
𝑥4 − 4𝑥2 + 5
𝑑𝑥
7. cos 3𝑥2
+ 1 𝑑𝑥
8.
3𝑐𝑜𝑠2
𝑥(𝑠𝑖𝑛 𝑥)
𝑥
𝑑𝑥
9.
𝑒2𝑥
1 + 𝑒4𝑥
𝑑𝑥
10. 6𝑡2
𝑒 𝑡3−2
𝑑𝑡

5. PENYELESAIAN
1. 3 𝑥3
− 5 4
𝑥2
𝑑𝑥
u = x3
– 5 du = 3x2
dx
= 𝑢4
𝑑𝑢
=
1
5
𝑢5
+ 𝑐
=
(𝑥3
− 5)5
5
+ 𝑐
2. 𝑒 𝑥4
𝑥3
𝑑𝑥
𝑢 = 𝑥4
= 𝑒 𝑥4 1
4
. 4𝑥3
𝑑𝑥
=
1
4
𝑒 𝑥3
4𝑥3
𝑑𝑥
=
1
4
𝑒 𝑢
𝑑𝑢
=
1
4
𝑒 𝑢
+ 𝑐
=
1
4
𝑒 𝑥4
+ 𝑐
3.
𝑡
𝑡2 + 7
𝑑𝑡
𝑢 = 𝑡2
+ 7 𝑑𝑢 = 2𝑡 𝑑𝑥
𝑡
𝑡2 + 7
𝑑𝑡

6. 1
2
2𝑡
𝑡2 + 7
𝑑𝑡
1
2
2𝑡
𝑡2 + 7
𝑑𝑡
1
2
𝑑𝑢
𝑢
1
2
𝐼𝑛 𝑢 + 𝑐
1
2
𝐼𝑛 𝑡2
+ 7 + 𝑐
4. 𝑥5
− 3𝑥
1
4 5𝑥4
− 3 𝑑𝑥
𝑢 = 𝑥5
− 3𝑥 𝑑𝑢 = 5𝑥4
− 3 𝑑𝑥
= 𝑢
1
4 𝑑𝑢
= 4𝑢
5
4 + 𝑐
= 4 𝑥5
− 3𝑥
5
4 + 𝑐
5.
𝑥3
− 2𝑥
𝑥4 − 4𝑥2 + 5 4
𝑑𝑥
𝑢 = 𝑥4
− 4𝑥2
+ 5 𝑑𝑢 = 4𝑥3
− 8𝑥 𝑑𝑥
=
1
4
.
4 𝑥3
− 2𝑥
𝑢4
𝑑𝑥
=
1
4
𝑑𝑢
𝑢4
=
1
4
𝐼𝑛 𝑢 + 𝑐
=
1
4
𝐼𝑛 𝑥4
− 4𝑥2
+ 5 + 𝑐

7. 6.
𝑥3
− 2𝑥
𝑥4 − 4𝑥2 + 5
𝑑𝑥
𝑢 = 𝑥4
− 4𝑥2
+ 5 𝑑𝑢 = 4𝑥3
− 8𝑥 𝑑𝑥
= 4 𝑥3
− 2𝑥
=
1
4
.
4(𝑥3
− 2𝑥)
𝑥4 − 4𝑥2 + 5
𝑑𝑥
=
1
4
𝑑𝑢
𝑢
=
1
4
𝐼𝑛 𝑢 + 𝑐
=
1
4
𝐼𝑛 𝑥4
− 4𝑥2
+ 5 + 𝑐
9.
𝑒2𝑥
1 + 𝑒4𝑥
𝑑𝑥
=
𝑒2𝑥
1 + 𝑒2𝑥(2)
𝑑𝑥
𝑢 = 1 + 𝑒2𝑥
𝑑𝑢 = 2. 𝑒2𝑥
𝑑𝑥
=
1
2
.
2. 𝑒2𝑥
1 + 𝑒2𝑥(2)
=
1
2
𝑑𝑢
𝑢
=
1
2
𝐼𝑛 𝑢 𝑑𝑥
=
1
2
𝐼𝑛 1 + 𝑒4𝑥
+ 𝑐

8. 10. 6𝑡2
𝑒 𝑡3−2
𝑑𝑡
𝑢 = 𝑡3
− 2 𝑑𝑢 = 3𝑡2
𝑑𝑡
= 6𝑡2
𝑒 𝑡3−2
𝑑𝑡
= 2 3𝑡2
𝑒 𝑡3−2
𝑑𝑡
=
1
3
. 3 2 . 3𝑡2
. 𝑒 𝑡3−2
𝑑𝑡
=
1
3
6 𝑑𝑢. 𝑒 𝑢
=
1
3
𝑒 𝑢
. 6 𝑑𝑢
=
1
3
𝑒 𝑡3−2
. 6 + 𝑐
= 2𝑒 𝑡3−2
+ 𝑐

9. Integrasi dengan bagian
Latihan 8.2
Gunakan integrasi dengan bagian untuk menemukan integral tak tentu yang paling umum.
1. 2𝑥.sin2x dx
2. 𝑥3
lnx dx
3. 𝑡𝑒 𝑡
dt
4. 𝑥 cos x dx
5. 𝑐𝑜𝑡−1
𝑥 𝑑𝑥
6. 𝑥2
𝑒 𝑥
𝑑𝑥
7. 𝑤( 𝑤 − 3)2
𝑑𝑤
8. 𝑥3
𝑖𝑛 4𝑥 𝑑𝑥
9. 𝑡 (𝑡 + 5)−4
𝑑𝑡
10. 𝑥 𝑥 + 2 . 𝑑𝑥

10. PENYELESAIAN
1. 2𝑥 sin 2𝑥 𝑑𝑥
Misalnya :
u = 2x du = x
dv = sin 2x dx v= sin 2𝑥𝑑𝑥 = -
1
2
cos2x
𝑢. 𝑑𝑣 = 𝑢𝑣 – 𝑢. 𝑑𝑢
2𝑥 sin 2𝑥 𝑑𝑥 = (2x) (-
1
2
cos 2x ) - (−
1
2
cos 2x ) . 2x
= -
2
2
cos 2x +
1
2
cos 2x dx
= - x cos 2x +
1
2
.
1
2
sin 2x
= - x cos 2x +
1
2
. sin 2x + c
2. 𝑥3
𝑖𝑛 𝑥 𝑑𝑥
Misalnya :
U= inx du =
1
𝑥
dx
dv= 𝑥3
dx v = 𝑥3
𝑑𝑥 =
𝑥4
4
𝑢. 𝑑𝑣 = 𝑢𝑣 – 𝑢. 𝑑𝑢
𝑥3
𝑖𝑛 𝑥 𝑑𝑥 = (in x) (
𝑥4
4
) -
𝑥4
4
.
1
𝑥
dx
=
𝑥4 𝑖𝑛𝑥
4
-
1
4
.
𝑥4
4
=
𝑥4 𝑖𝑛𝑥
4
-
𝑥4
16
+ c
3. 𝑡𝑒 𝑡
𝑑𝑡
Misalnya :
U = t du = dt
dv = 𝑒 𝑡
dt v = 𝑒 𝑡
dt = 𝑒 𝑡
𝑢. 𝑑𝑣 = 𝑢. 𝑣 – 𝑢. 𝑑𝑢

11. 𝑡𝑒 𝑡
𝑑𝑡 = (t) (𝑒 𝑡
) - 𝑒 𝑡
dt
= 𝑡𝑒 𝑡
- 𝑒 𝑡
dt
= 𝑡𝑒 𝑡
- 𝑒 𝑡
+ c
4. 𝑥 cos 𝑥 𝑑𝑥
Misalnya :
U= x du = dx
dv = cos x dx v = cos 𝑥 𝑑𝑥 = sin x
𝑢. 𝑑𝑣 = 𝑢. 𝑣 – 𝑢. 𝑑𝑢
𝑥 cos 𝑥 𝑑𝑥 = ( x ) ( sin x ) - sin 𝑥 𝑑𝑥
= sin x + cosx dx
= sin x + cosx + c
5. 𝑐𝑜𝑡−1
( x ) dx
Misalnya :
U = sin𝑥−1
Du= cos𝑥−1
Subtitusi du = sin𝑥−1
du = cos𝑥−1
𝑐𝑜𝑠𝑥 −1
𝑠𝑖𝑛𝑥 −1 dx =
𝑑𝑢
𝑢
Salve integral
= in (u) + c
Subsitusi kembali
U=sin𝑥−1
= in (sin𝑥−1
) + 𝑐
6. 𝑥2
𝑒 𝑥
𝑑𝑥
Misalnya :
U = 𝑥2
du = 2x
dv = 𝑒 𝑥
dx v = 𝑒 𝑥
dx = 𝑒 𝑥
𝑢. 𝑑𝑣 = u.v - 𝑢.du
𝑥2
𝑒 𝑥
𝑑𝑥 = 𝑥2
𝑒 𝑥
- 𝑥
2
. 2𝑥
=𝑥𝑒2𝑥
- 2𝑥. 𝑑𝑥
=𝑥𝑒2𝑥
- x+c
7. 𝑤(𝑤 − 3)2
𝑑𝑤
Misalnya :
U= w du= dw

12. dv = (𝑤 − 3)2
𝑑𝑤 𝑣 = 2𝑤 − 6 = 𝑤 − 3
𝑢. 𝑑𝑣 = u.v - 𝑢.du
𝑤(𝑤 − 3)2
𝑑𝑤 = 𝑤. 𝑤 − 3 − 𝑤. 𝑑𝑤
= 𝑤2
− 3𝑤 −
1
2
𝑤 + 𝑐
8. 𝑥3
𝑖𝑛 4𝑥 𝑑𝑥
Misalnya :
U= in4x du=
1
4𝑥
𝑑𝑥
dv= 𝑥3
𝑑𝑥 v = 𝑥3
dx =
1
4
𝑥4
𝑢. 𝑑𝑣 = u.v - 𝑣.du
𝑥3
𝑖𝑛 4𝑥 𝑑𝑥 = in4x.
1
4
𝑥4
- in4x .
1
4𝑥
𝑑𝑥
=
1
4
𝑥4
𝑖𝑛4𝑥 −
1
5
𝑥5
∶
1
2
16𝑥2
+ 𝑐
=
1
4
𝑥4
𝑖𝑛4𝑥 -
2𝑥5
80𝑥2 + c
9. 𝑡(𝑡 + 5)−4
𝑑𝑡
Misalnya :
U= t du= dt
dv =(𝑡 + 5)−4
𝑣 = −4𝑡−3
− 20−3
= 2𝑡−2
+ 10−2
𝑢. 𝑑𝑣 = u.v - 𝑣.du
𝑡(𝑡 + 5)−4
𝑑𝑡 =( t. 2𝑡−2
+ 10−2
) - 2𝑡−2
+ 10−2
. 𝑑𝑡
= 20𝑡−4
+ (2𝑡 + 10 + 𝑑𝑡
10. 𝑥 𝑥 + 2 .dx
Misalnya :
U = x du = dx
Dv= 𝑥 + 2 dx v= (𝑥 + 2)
1
2 =2𝑥1
1
2 +0.671
1
2
𝑢. 𝑑𝑣 = u.v - 𝑣.du
𝑥 𝑥 + 2 .dx = x . 2𝑥1
1
2 +0.671
1
2 - 2𝑥1
1
2 + 0.671
1
2 . dx
= x.2,67𝑥
3
2 - (2𝑥
3
2 + 0,67
3
2) dx
= 2,67𝑥2
3
2 - 2,67𝑥
6
2 + c

13. Integrasi dengan menggunakan tabel rumus
terpisahkan
Latihan 8.3
Gunakan tabel rumus integral dalam Lampiran C untuk menemukan integral tak tentu yang
paling umum.
1. cot 𝑥 𝑑𝑥
2.
1
𝑥+2 (2𝑥+5)
𝑑𝑥
3. 𝑙𝑛𝑥 2
𝑑𝑥
4. 𝑥 cos 𝑥 𝑑𝑥
5.
𝑥
𝑥+2 2 𝑑𝑥
6. 3𝑥𝑒 𝑥
𝑑𝑥
7. 10 𝑤 + 3 𝑑𝑤
8. 𝑡(𝑡 + 5)−1
𝑑𝑡
9. 𝑥 𝑥 + 2 𝑑𝑥
10.
1
sin 𝑢 cos 𝑢
𝑑𝑢

14. PENYELESAIAN
1. cot 𝑥 𝑑𝑥
( Formula nomor 7)
Penyelesaian :
𝑐𝑜𝑡 𝑥 𝑑𝑥 =
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥
𝑑𝑥
Misalkan :
𝑢 = sin 𝑥
𝑑𝑢 = cos 𝑥 𝑑𝑥
Subsitusi 𝑑𝑢 = cos 𝑥, 𝑈 = sin 𝑥
cos 𝑥
sin 𝑥
𝑑𝑥 =
𝑑𝑢
𝑢
𝑠𝑎𝑙𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
ln 𝑢 + 𝐶
subsitusi kembali 𝑈 = sin 𝑥
𝑙𝑛 sin 𝑥 + 𝑐
2.
1
𝑥+2 (2𝑥+5)
𝑑𝑥
=
1
𝑥 + 2 (2𝑥 + 5)
=
𝐴
𝑥 + 2
+
𝐴
2𝑥 + 5
𝐴 =
1
𝑥 + 2 (2.2 + 5)
=
1
9
𝐵 =
1
5 + 2 (2𝑥 + 5)
=
1
7
Sehingga :
1
𝑥 + 2 2𝑥 + 5
𝑑𝑥 =
1
𝑥 + 2 2𝑥 + 5

15. =
1
9
𝑥 + 2
𝑑𝑥 +
1
9
2𝑥 + 5
𝑑𝑥
=
1
9
𝑙𝑛 𝑥 + 2 +
1
7
ln 2𝑥 + 5 + c
3. 𝑙𝑛𝑥 2
𝑑𝑥 = 𝑙𝑛𝑥 𝑙𝑛𝑥 𝑑𝑥
Missal :
U = ln x 𝑑𝑢 = (
1
𝑥
)2
Dv = dx
dv = 𝑑𝑥
v = x
(𝑙𝑛𝑥)2
𝑑𝑥 = 𝑢𝑣 − 𝑣𝑑𝑢 (x ln )
= (𝑙𝑛𝑥)2
. x - 𝑥
1
𝑥2 𝑑𝑥
= 𝑥. (𝑙𝑛𝑥)2
-
1
𝑥
𝑑𝑥
= 𝑥. (𝑙𝑛𝑥)2
x - 𝑥−1
𝑑𝑥
= 𝑥. (𝑙𝑛𝑥)2
-
1
0
𝑥0
+ 𝑐
= 𝑥. (𝑙𝑛𝑥)2
- ~ + 𝑐
= ln x ( x ln x-x ) – (𝑥 ln 𝑥 − 𝑥) .
1
𝑥
=x (ln x)2
- x ln x -
4. 𝑥 cos 𝑥 𝑑𝑥
Penyelesaian :
𝑈 = 𝑋 → 𝑑𝑢 = 𝑑𝑥
𝑑𝑣 = 𝑐𝑜𝑠𝑥 → 𝑣 = 𝑠𝑖𝑛𝑥
𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢

16. 𝑥𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥
𝑥𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝑐
5.
𝑥
𝑥+2 2 𝑑𝑥
Penyelesaian :
𝑥
𝑥+2 2 =
𝐴
𝑥+2
+
𝐵
𝑥+2
=
𝐴 𝑥+2 +𝐵
𝑥+2
2
𝐴 = 2
𝐴 + 𝐵 = 0 = −2
Sehingga :
𝑥
𝑥 + 2 2
𝑑𝑥 =
𝑑𝑥
𝑥 + 2
–
𝑑𝑥
𝑥 + 2 2
𝑀𝑖𝑠𝑎𝑙𝑙 𝑢 = 𝑥 + 2 → 𝑑𝑢 = 𝑑𝑥
𝑑𝑥
𝑥 + 2
–
𝑑𝑥
𝑥 + 2 2
=
𝑑𝑢
𝑢
–
𝑑𝑢
𝑢2
= 2𝑙𝑛 +
2
𝑢
+ 𝑐
2𝑙𝑛 𝑥 + 2 +
2
𝑥+2
+ 𝑐
6. 3𝑥𝑒 𝑥
𝑑𝑥
U = 3x dv = 𝑒 𝑥
𝑑𝑥
𝑑𝑢
𝑑𝑥
= 3 v = 𝑒 𝑥
𝑑𝑥 = 𝑒 𝑥
du = 3 dx
𝑢𝑑𝑣 = u.v – 𝑣 𝑑𝑢

17. = (3x) . (𝑒 𝑥
) – 𝑒 𝑥
. 3 𝑑𝑥
= 3x 𝑒 𝑥
− 3𝑒 𝑥
7. 10 𝑤 + 3 dw
( Formula nomor 2)
10 𝑤 + 3 dw = (10 𝑤 + 3)
1
2 dw
=
1
1
2
+ 1
(10 𝑤 + 3)
1
2
+1
+ 𝑐
=
2
3
(10 𝑤 + 3)
3
2 + 𝑐
8. 𝑡(𝑡 + 5)−1
𝑑𝑡
=
𝑡
𝑡+5
dt = 𝑡 (𝑡 + 5)−1
𝑑𝑡
Missal:
U = t + 5 U= t+5
𝑑𝑢
𝑑𝑡
= 1 t = (u-5)
𝑑𝑢 = 𝑑𝑡 t=u→u=t+5 =5
t = 2 → u=t+5 = 7
=
𝑡
𝑡+5
dt = 𝑡 (𝑡 + 5)−1
𝑑𝑡 = 𝑢 − 5 𝑢−1
𝑑𝑢 = 𝑢0
− 5𝑢−1
𝑑𝑢
(𝑢0
− 5𝑢) … … … … . = 𝑢 − 𝑢
−5𝑢−1
+1 du
−5(𝑢1
−
1
5
𝑥 ) 𝑑𝑥
-5 (ln 𝑢 -
1
5
0+1
𝑥0+1
)
-5 ( ln 𝑡 + 5 -
1
5
x)

18. -5 ln 𝑡 + 5 + x
9. 𝑥 𝑥 + 2 𝑑𝑥
𝑚𝑖𝑠𝑎𝑙 𝑢 = 𝑥 + 2 → 𝑥 = 𝑢 − 2
𝑑𝑢 = 𝑑𝑥
Sehingga integral diatas dapat menjadi :
= 𝑖𝑛𝑡 𝑢 − 2 𝑈 𝑑𝑢
= 𝑖𝑛𝑡 𝑢 − 2 𝑈
1
2 𝑑𝑢
= 𝑖𝑛𝑡 𝑈
5
2 − 𝑈
1
2 𝑑𝑢
=
2
7
𝑈
2
7 −
2
3
𝑈
3
2 + 𝐶
= 𝑖𝑛𝑡 (𝑥 + 2)
5
2 −
2
3
(𝑥 + 2)
3
2 + 𝐶
10.
1
sin 𝑢 cos 𝑢
𝑑𝑥 = (sin 𝑢)−1
(cos 𝑢)−1
dx