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

Cara cepat yang sesungguhnya hanya dapat diperoleh di NICEinstitute.
Hanya 2 baris loh.

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 Tanzalin 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

Cara Tanzalin ini bukanlah cara cepat.
Cara cepat yang sesungguhnya hanya dapat diperoleh di NICEinstitute.
Hanya 2 baris loh.

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 Tanzalin 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
  |  (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


Cara Tanzalin ini bukanlah cara cepat.
Cara cepat yang sesungguhnya hanya dapat diperoleh di NICEinstitute
Hanya 2 baris loh.