Page 43 - TUTORIAL_MAT_IPS_K11-2

INTEGRAL FUNGSI ALJABAR

1 3 1
4
5
4
5
4
5
10. ∫ 3 √2 − 1 = ∫ (2 − 1)4 = 1 (2 − 1)4 +1 + =
2.5.(1+ )
4
6 5
5
(2 − 1)4 + (cara cepat).
25

1
3 2 − 1 3 2 − +1 9 2 2
11. ∫ 3 = ∫ 3 ( + 1) 3 = 1 ( + 1) 3 + = ( + 1)3 +
2
√ +1 1.2.(1+(− )) 4
3
(cara cepat).

3
(√ +3)
12. ∫ =
√
′
Misal : = √ + 3 → = 1 → = 1 → = 2√
2√ 2√
3
(√ +3) 3 1 1 4
4
3
∫ = ∫ 2√ = ∫ = + = (√ + 3) +
√ √ 4 4

−2
1 1
13. ∫ (4 + ) ( ) =
2
1 1 1
′
2
Misal : = 4 + → = − → = − → = −
2 2
−2 −1
1 1 1 1
2
∫ (4 + ) ( ) = ∫ −2 . − = − ∫ −2 = −1 + = (4 + ) +
2 2

14. ∫ √2 − 1 =
1
1
2
2
Misal : = √2 − 1 → = 2 − 1 → ( + 1) = → . 2 = → =
2 2
1 2 1 4 1 2 1 5 1 3
∫ √2 − 1 = ∫ ( + 1). . = ∫ + = + + =
2 2 2 10 6
1 5 1 3
(2 − 1)2 + (2 − 1)2 +
10 6
*Selain menggunakan cara substitusi, soal ini juga bisa diselesaikan menggunakan cara
integral parsial seperti soal no. 16

2
2
2
+4 −1 −1+4 −1 4 4
15. ∫ = ∫ = ∫ + ∫ = ∫ 1 + ∫ =
2
2
2
2
2
−1 −1 −1 −1 −1
∫ 1 = +
4 2 −1 4 2 2
∫ = ∫ 4 ( − 1) = ln( − 1) + = 2 ln( − 1) +
2
−1 1.2
2
+4 −1
2
∫ = + 2 ln( − 1) +
2
−1

‘LEARNING IS FUN’ 42