INTEGRAL PARSIAL
Bentuk umum:∫ f(x) ∙ gⁿ (x) dx
Rumus:
∫ u ∙ dv = u ∙ v - ∫ v ∙ du
Contoh 1:∫ 4x (2x + 6)⁵ dx
Jawab:misalkan
u = 4x ⇒ du = 4 dx
dv = (2x + 6)⁵ dx ⇒ v = ∫
(2x + 6)⁵ dx = ⅙ ∙ ½ ∙
(2x + 6)⁶ + c
∫ 4x (2x + 6)⁵ dx
= 4x ∙ ⅙ ∙ ½ ∙ (2x + 6)⁶ - ∫ ⅙ ∙ ½ ∙ (2x + 6)⁶ ∙ 4 dx
= ⅓ (2x + 6)⁶ - ⅓ ∙ ½ ∙ (1/7) (2x + 6)⁷ + c
= ⅓ (2x + 6)⁶ - (1/42) (2x + 6)⁷ + c
Contoh 2:
Hasil dari ∫ 8x² cos 2x dx = ….
Jawab:
∫ 8x² cos 2x dx
u = 8x² ⇒ du = 16x dx
dv = cos 2x dx ⇒ v = ∫
cos 2x dx = ½ ∙
sin 2x
∫ 8x² cos 2x dx;
RUMUS: ∫ u dv = u ∙ v – ∫ v du
= 8x² ∙ ½ ∙ sin 2x – ∫ ½ ∙ sin 2x ∙16x dx
= 4x² ∙ sin 2x – 8 ∫
x ∙ sin 2x dx
∫ x ∙ sin 2x dx = ....
u = x ⇒ du = dx
dv = sin 2x dx ⇒ v = ∫
sin 2x dx = –½ ∙
cos 2x
∫ x ∙ sin 2x dx
= x (–½ ∙ cos 2x) – ∫ –½ ∙ cos 2x dx
= –½x ∙ cos 2x + ¼ ∙ sin 2x + c
4x² ∙ sin 2x – 8 ∫ x ∙ sin 2x dx
= 4x² ∙ sin 2x – 8(–½x ∙ cos 2x + ¼ ∙ sin 2x) + c
= 4x² ∙ sin 2x + 4x ∙ cos 2x – 2 ∙ sin 2x + c
Dapat pula menggunakan cara Tabulasi sebagai berikut:
8x² (turunkan) | cos
2x (integralkan)
16x | ½∙sin 2x
16 | –¼∙cos 2x
0 | –⅛∙sin 2x
∫ 8x² cos 2x dx
= 8x²(½ ∙ sin 2x) – 16x(–¼ ∙ cos 2x) + 16(–⅛ ∙ sin 2x)
= 4x² ∙ sin 2x + 4x ∙ cos 2x – 2 ∙ sin 2x + c
Contoh 4:
∫ 4x³ (2x – 4)⁵ dx = ….
Jawab:
∫ 4x³ (2x - 4)⁵ dx
u = 4x³ ⇒ du = 12x² dx
dv = (2x – 4)⁵ dx ⇒ v = ∫
(2x – 4)⁵ dx
= (1/12)(2x – 4)⁶ +
c
∫ 4x³ (2x – 4)⁵ dx
= 4x³(1/12)(2x – 4)⁶ – ∫ (1/12)(2x – 4)⁶ ∙ 12x² dx
= ⅓ x³(2x – 4)⁶ – ∫x² ∙ (2x – 4)⁶ dx
∫x² ∙ (2x – 4)⁶ dx
u = x² ⇒ du = 2x dx
dv = (2x – 4)⁶ dx ⇒ v = ∫
(2x – 4)⁶ dx
= (1/14)(2x – 4 )⁷ +
c
∫x² ∙ (2x – 4)⁶ dx
= x²(1/14)(2x – 4 )⁷ – ∫ (1/14)(2x – 4 )⁷ ∙ 2x dx
= (1/14)x²(2x – 4)⁷ – (1/7) ∫ x ∙ (2x – 4)⁷ dx
∫ x ∙ (2x – 4)⁷ dx
u = x ⇒ du = dx
dv = (2x – 4)⁷ dx ⇒ v = ∫
(2x – 4)⁷ dx
= (1/16)(2x – 4 )⁸ +
c
∫ x ∙ (2x – 4 )⁷ dx
= x(1/16)(2x – 4 )⁸ – ∫ (1/16)(2x – 4 )⁸ dx
= (1/16)x(2x – 4 )⁸ – (1/16) ∫ (2x – 4 )⁸ dx
= (1/16)x(2x – 4 )⁸ – (1/16)(1/18)(2x – 4 )⁹
= (1/16)x(2x – 4 )⁸ – (1/288)(2x – 4 )⁹
∫ 4x³ (2x – 4)⁵ dx
= 4x³(1/12)(2x – 4)⁶ – ∫ (1/12)(2x – 4)⁶∙ 12x² dx
= ⅓ x³(2x – 4)⁶ – ∫x² ∙ (2x – 4)⁶ dx
= ⅓ x³(2x – 4)⁶ – {(1/14)x²(2x – 4 )⁷ – (1/7) ∫ x ∙ (2x – 4
)⁷ dx}
= ⅓ x³(2x – 4)⁶ – (1/14)x²(2x – 4 )⁷ + (1/7) ∫ x ∙ (2x – 4)⁷
dx
= ⅓ x³(2x – 4)⁶ – (1/14)x²(2x – 4)⁷ + (1/7){(1/16)x(2x – 4)⁸
– (1/288)(2x – 4)⁹}= ⅓ x³(2x – 4)⁶ – (1/14)x²(2x – 4)⁷ + (1/112)x(2x – 4)⁸ –
(1/2016)(2x – 4)⁹ + c
Dapat pula menggunakan cara Tabulasi sebagai berikut:
4x³ (turunkan) | (2x
– 4)⁵ (integralkan)
12x² | (1/12)(2x – 4)⁶
24x | (1/12)(1/14)(2x – 4)⁷
24 | (1/12)(1/14)(1/16)(2x – 4)⁸
0 |
(1/12)(1/14)(1/16)(1/18)(2x – 4)⁹
∫ 4x³ (2x – 4)⁵ dx
= 4x³(1/12)(2x – 4)⁶ – 12x²(1/12)(1/14)(2x – 4)⁷ + 24x
(1/12)(1/14)(1/16)(2x – 4)⁸
–
24(1/12)(1/14)(1/16)(1/18)(2x – 4)⁹ + c
= ⅓ x³(2x – 4)⁶ – (1/14)x²(2x – 4)⁷ + (1/112)x(2x – 4)⁸ –
(1/2016)(2x – 4)⁹ + c
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