tolong bantuan nya ngerjain nya please

integral

1. ∫(x² - 2x + 5√x)√x dx
= ∫ (x²√x - 2x√x + 5x) dx
= ∫ x⁵/² - 2 x³/² + 5x dx
= ²/₇ x⁷/² - 2(²/₅) x⁵/² + ⁵/₂ x² + c
= ²/₇ x³√x - ⁴/₅ x²√x +⁵/₂ x² + c

2.
∫ 8 cos 4x sin 2x dx =
= 4 ∫ 2 cos 4x sin 2x dx
= 4 ∫ sin ¹/₂(4x + 2x) - sin ¹/₂ (4x - 2x) dx
= 4 ∫ sin 3x - sin x  dx
= 4 [ - 1/3 cos 3x  - (-cos x) ] +c
= -4/3 cos 3x + 4 cos x + c

3.
₁ᵃ∫ (2x -3) = 6
= [x² - 3x ]ᵃ₁ = 6  dan a > 1
(a² - 3a )-(1² -3(1)) = 6
a² - 3a - (1-3)  = 6
a² - 3a + 2 - 6 = 0
a² - 3a - 4 = 0
(a - 4)(a + 1)= 0
a = 4 atau a = - 1 , dan a > 1
hp a = 4

4.
∫ (x²-3) √(x³ -9x +6)³ dx =
u = x³ -9x + 6
du = 3x² - 9 dx
du = 3(x²-3) dx
(x² - 3) dx = 1/3 du
∫(x²-3) √(x³-9x+6)³ dx =  ∫(¹/₃ du)(u³/²) du
= 1/3 . (2/5) (u)⁵/2 + c
= 2/15 u√u + c
= 2/15 (x³-9x +6) √(x³ -9x + 6) + c

5. dengan parsial (tabel )
∫(4x² -6) sin 2x dx =

turunan .................integral
4x² - 6  ..................sin 2x
8x ......................-1/2 cos 2x
8 .......................- 1/4 sin 2x
0 .......................1/8 cos 2x
=
∫(4x² -6) sin 2x dx =
= (4x²-6)(-1/2) cos 2x  - (8x)(-1/4) sin 2x +8(1/8) cos 2x + c
= (3-2x²) cos 2x + 2x sin 2x + cos 2x + c
..