Soal Integral 1

Contoh 1:
∫ {(3x³ – 4x² + 3x) /(x² + 1)} dx = ….

Jawab:










Contoh 2:
  ∫ √(1 + sin x) dx = ....
 
Jawab:
∫ √(1 + sin x) dx
= ∫ √{1 + cos (½π - x} dx
= ∫ √{1 + cos (½π - x)} dx
= ∫ √{2 cos² (¼π - ½x)} dx
= √2 ∫ cos (¼π - ½x)} dx
= -2√2 sin (¼π - ½x) + c.

Contoh 3:  

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.

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
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


Cara Tanzalin ini bukanlah cara cepat.
Cara cepat yang sesungguhnya hanya dapat diperoleh di NICEinstitute.