Berikut ini beberapa contoh soal integral tak tentu fungsi aljabar dan penyelesaiannya.
Soal: Hasil dari ∫(4x³ + 15x² – 4x – 16) dx adalah…
Jawaban:
∫(4x³ + 15x² – 4x – 16) dx
= (4/4)x⁴ + (15/3)x³ – (4/2)x² – 16x + C
= x⁴ + 5x³ – 2x² – 16x + C
Soal: Hasil dari ∫(2x-1)(4x² – x – 1) dx adalah…
Jawaban:
∫(2x-1)(4x² – x – 1) dx
= ∫(8x³ – 2x² – 2x – 4x² + x + 1) dx
= ∫ (8x³ – 6x² – x + 1) dx
= (8/4)x⁴ – (6/3)x³ – (1/2)x² + x + C
= 2x⁴ – 2x³ – (1/2)x² + x + C
Soal: Hasil dari ∫2x² (x – 3) dx adalah…
Jawaban:
∫2x² (x – 3) dx
= ∫ (2x³ – 6x²) dx
= (2/4)x⁴ – (6/3)x³ + C
= (1/2)x⁴ – 2x³ + C
Soal: Hasil dari ∫(x-2)(x+1)² dx adalah…
Jawaban:
∫(x-2)(x+1)² dx
= ∫(x-2)(x+1)(x+1) dx
= ∫(x-2)(x² + 2x +1) dx
= ∫(x³ + 2x² + x – 2x² – 4x – 2) dx
= ∫(x³ – 3x -2) dx
= (1/4)x⁴ – (3/2)x² – 2x + C
Soal: Hasil dari ∫(6x+15)(x² + 5x – 2)⁵ dx adalah…
Jawaban:
Misalkan:
u = x² + 5x -2
du = (2x + 5) dx
Maka:
∫(6x+15)(x² + 5x – 2)⁵ dx
= ∫(x² + 5x – 2)⁵ (6x+15) dx
= ∫(x² + 5x – 2)⁵ (3 ⨯(2x+5)) dx
= ∫u⁵ 3du
= ∫3u⁵ du
= (3/6)u⁶ + C
= (1/2)u⁶ + C
= (1/2)(x² + 5x – 2)⁶ + C
Soal: Hasil dari ∫2x² (x³+2)⁵ dx adalah…
Jawaban:
Misalkan
u = x³ + 2
du = 3x² dx
du/3 = x² dx
Maka:
∫2x² (x³+2)⁵ dx
= ∫(x³ + 2)⁵ 2x² dx
= ∫ u⁵ 2 (du/3)
= (2/3) ∫u⁵ du
= (2/3) (1/6) u⁶ + C
= (2/18) u⁶ + C
= (1/9) (x³ + 2)⁶ + C
Soal: Hasil dari ∫(x² +2) √(x³+6x+2) dx adalah…
Jawaban:
Misalkan:
u=x³+6x+1
du=3x²+6 dx
du=3(x²+2) dx
du/3 = (x²+2) dx
Maka
∫(x² +2) √(x³+6x+2) dx = ∫√(x³+6x+1) (x²+2)dx
= ∫√(u ) (du/3)
= (1/3) ∫√(u) du
= (1/3) ∫ \(u^{1/2}\) du
= (1/3) (1/(3/2)) \(u^{3/2}\) + C
= (1/3) (2/3) \(u^{3/2}\) + C
= (2/9) \(u^{3/2}\) + C
= (2/9) \((x³+6x+1)^{3/2}\) + C
⁼ (2/9) (x³+6x+2)√(x³+6x+2) + C
Soal: Hasil dari ∫(2x√x + 1/x) (x – 4/(x√x))dx adalah…
Jawaban:
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Soal: Hasil dari ∫(x-1)/(1+2x-x²)³ dx adalah…
Jawaban:
Pelajari Juga: Rangkuman dan Contoh Integral Tak Tentu dan Integral Tentu
Demikian pembahasan tentang “Contoh Soal Integral Tak Tentu Fungsi Aljabar SMA“. Semoga Bermanfaat.
