1. integral 6 per akar 4-16x^2

Nomor 1
∫ 6 dx / √(4 – 16x²)
= ∫ 3 dx / √(1 – 4x²) = 3 ∫ dx / √(1 – 4x²)
∫ du / √(a² – u²) = arcsin (u / a) + C
3 ∫ dx / √(1 – 4x²) = …
u = 2x
du / dx = 2
= 3 [1/2 ∫ du / √(1 – u²)]
= 3/2 arcsin u + C
= 3/2 arcsin (2x) + C

Nomor 2
∫ 10 dx / (x² – 6x + 34)
= 10 ∫ dx / (x² – 6x + 34) = 10 ∫ dx / [(x – 3)² + 25]
u = (x – 3) / 5 → du / dx = 1 / 5
= 10 ∫ 5 du / [(5u)² + 25]
= 10 ∫ 1/5 du / (u² + 1)
= 2 ∫ du / (u² + 1)
= 2 arctan u + C
= 2 arctan [(x – 3) / 5] + C

Nomor 3
∫ (x – 1) dx / (x – 2)² 
(x – 1) / (x – 2)² = A / (x – 2) + B / (x – 2)²
(x – 1)(x – 2)² / (x – 2)² = A(x – 2)² / (x – 2) + B(x – 2)² / (x – 2)²
x – 1 = A(x – 2) + B
Untuk akar penyebut sama dengan 2
2 – 1 = A(2 – 2) + B → B = 1

x – 1 = Ax – 2A + 1
x – 1 = Ax + (-2A + 1)
-2A + 1 = -1 → A = 1
Jadi
(x – 1) / (x – 2)² = 1 / (x – 2) + 1 / (x – 2)²
∫ [1 / (x – 2) + 1 / (x – 2)²] dx
= ㏑ |x – 2| – 1 / (x – 1) + C