Page 44 - TUTORIAL MATEMATIKA IPA SEMESTER 2

INTEGRAL FUNGSI ALJABAR

16. ∫ √4 + 1 =

Turunan Integral Tanda
x 1
(4 + 1)2
1 1 2 3 1 3 +
. (4 + 1)2 = (4 + 1)2
4 3 6
5
0 1 1 2 5 1 (4 + 1)2 -
. . (4 + 1)2 =
6 4 5 60

1 3 1 5 1 3 1 5
∫ √4 + 1 = . (4 + 1)2 − 1. (4 + 1)2 + = (4 + 1)2 − (4 + 1)2 +
6 60 6 60

3
17. ∫ √3 − 2 =

Turunan Integral Tanda
3
1
(3 − 2 )2
2
3 1 2 3 1 3 +
− . (3 − 2 )2 = − (3 − 2 )2
2 3 3
5
5
6x − . − . (3 − 2 )2 = 1 (3 − 2 )2 -
1
1 2
3 2 5 15
7
7
1 2
6 1 . − . (3 − 2 )2 = − 1 (3 − 2 )2 +
15 2 7 105
0 1 1 2 9 1 9 -
− . − . (3 − 2 )2 = (3 − 2 )2
105 2 9 945

3
∫ √3 − 2 =
3
9
5
7
1
3
2
. − (4 + 1)2 − 3 . 1 (3 − 2 )2 − 6 . 1 (3 − 2 )2 − 6. 1 (3 − 2 )2 + =
3 15 105 945
1 3 3 1 2 5 2 7 2 9
− (4 + 1)2 − (3 − 2 )2 − (3 − 2 )2 − (3 − 2 )2 +
3 5 35 315

4 +1 4 +1
18. ∫ = ∫ =
2
3 +2 −1 (3 −1)( +1)
4 +1 ( +1)+ (3 −1) ( +3 ) +( − )
= + = = →
(3 −1)( +1) 3 −1 +1 (3 −1)( +1) (3 −1)( +1)
7 3
+ 3 = 4 ; − = 1 , eliminasi kedua persamaan ini didapatkan = , =
4 4
7 3
4 +1 ⁄ ⁄ 7 1 3 1
∫ = ∫ 4 + 4 = . . ln|3 − 1| + . . ln|3 − 1| + =
2
3 +2 −1 3 −1 −1 4 3 4 3
7 1
ln|3 − 1| + ln|3 − 1| +
12 4

√ + −√ + √ + −√ + √ + −√ +
19. ∫ = ∫ . = ∫ = ∫ =
√ + +√ + √ + +√ + √ + −√ + ( + )−( + ) −
√ + −√ + 1 1 +1 1 1 +1 2 3 2 3
∫ = 1 ( + )2 − 1 ( + )2 + = ( + )2 − ( + )2 +
1 1+ 1+ 3 3
2 2

‘LEARNING IS FUN’ 43