Maths A Level Question - C3 Integration?

By 'inspection' you say?? When integrating a function using inspection, we are looking for a function that 'differentiates' to give us the existing function.

Inspecting the right bracket we see that if we differentiate it we will get the first bracket.

so, we try y = (x³ - x + 4)⁵

Hence, dy/dx = 5(3x² - 1)(x³ - x + 4)⁴

=> ∫ 5(3x² - 1)(x³ - x + 4)⁴ dx = (x³ - x + 4)⁵

i.e. ∫ (3x² - 1)(x³ - x + 4)⁴ dx = (1/5)(x³ - x + 4)⁵ + C
------------------------------------------------------------

With the next, we'll try y = (x² - 2x + 4)⁸

so, dy/dx = 8(2x - 2)(x² - 2x + 4)⁷

Hence, ∫ 8(2x - 2)(x² - 2x + 4)⁷ dx => (x² - 2x + 4)⁸

or, ∫ 16(x - 1)(x² - 2x + 4)⁷ dx => (x² - 2x + 4)⁸

i.e. ∫ (x - 1)(x² - 2x + 4)⁷ dx => (1/16)(x² - 2x + 4)⁸ + C

:)>

14)
Let u = x³ - x + 4
Then, du = (3x² - 1) dx

∫(3x² - 1) (x³ - x + 4)⁴ dx
= ∫(x³ - x + 4)⁴ [(3x² - 1) dx]
= ∫u⁴ du
= (1/5) u⁵ + C
= (1/5) (x³ - x + 4)⁵ + C


====
16)
Let u = x² - 2x + 4
Then, du = 2x - 2

∫(x - 1) (x² - 2x + 4)⁷ dx
= (1/2) ∫(2x - 2) (x² - 2x + 4)⁷ dx
= (1/2) ∫(x² - 2x + 4)⁷ [(2x - 2) dx]
= (1/2) ∫u⁷ du
= (1/2) * (1/8) u⁸ + C
= (1/16) (x² - 2x + 4)⁸ + C

14) int [3(x^2) - 1]*{[(x^3) - x + 4]^4} dx

u = (x^3) - x + 4; du = [3(x^2) - 1] dx

int u^4 du = (u^5)/5 = (1/5)*[((x^3) - x + 4)^5]

16) int (x - 1)*[((x^2) - 2x + 4)^7] dx

u = (x^2) - 2x; du = (2x - 2) dx; (1/2)*du = (x - 1) dx

int (1/2)*(u^7) du = (u^8)/16 = (1/16)*[((x^2) - 2x + 4)^8]

..
Substitution

14)
∫ (3x² - 1)(x³ - x + 4)⁴ dx

let u = x³ - x + 4
du/dx = 3x² - 1
dx = du/(3x² - 1)

∫ (3x² - 1)(x³ - x + 4)⁴ dx
= ∫ (3x² - 1)u⁴ dx
= ∫ (3x² - 1)u⁴ * du/(3x² - 1)
= ∫ u⁴ du
= ⅕ u⁵ + C
= ⅕ ( x³ - x + 4 )⁵ + C
┄┄┄┄┄┄┄┄┄┄

16)
∫( x -1)(x² - 2x + 4)⁷ dx

let u = x² - 2x + 4
du/dx = 2x - 2
du/dx = 2(x - 1)
dx = du/[2(x - 1)]

∫( x -1)(x² - 2x + 4)⁷ dx
= ∫( x -1)u⁷ dx
= ∫( x -1)u⁷ * du/[2(x - 1)]
= ∫ ½ u⁷ du
= (1/16) u⁸ + C
= (1/16) ( x² - 2x + 4 )⁸ + C
┄┄┄┄┄┄┄┄┄┄┄