Evaluate definite integral ∫[(x-x^3)^(1/3)]/x^4 dx from 1/3 to 1?

I = ∫ [ ( x - x³ )¹ʹ³ / x⁴ ] dx ... on [ 1/3, 1 ]

I = ∫ { x³ [ (x/x³) - 1 ] }¹ʹ³ / x⁴ dx

I = ∫ { [ x ( xֿ² - 1 )¹ʹ³ ] / x⁴ } dx

I = ∫ (xֿ² - 1)¹ʹ³ · xֿ³ dx ............................ (1)

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Let : u = xֿ² - 1.

Then : du/dx = -2.xֿ³

i.e., : xֿ³ dx = (-1/2) du.

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Then, from (1),

I = ∫ u¹ʹ³ · (-1/2) du

I = (-1/2) · [ ( u⁴ʹ³ ) / (4/3) ]

I = (-3/8) [ (xֿ² - 1)⁴ʹ³ ] .............. on [ 1/3, 1 ]

I = (-3/8) [ (1-1)⁴ʹ³ - (9-1)⁴ʹ³ ]

I = (-3/8) [ 0 - (2³)⁴ʹ³ ]

I = (-3/8)( -16)

I = 6 ................................................ Ans.

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Happy To Help !

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x = sin(t)

dx = cos(t) * dt

(sin(t) - sin(t)^3)^(1/3) * cos(t) * dt / sin(t)^4 =>

(sin(t) * (1 - sin(t)^2))^(1/3) * cos(t) * dt / sin(t)^4 =>

(sin(t) * cos(t)^2)^(1/3) * cos(t) * dt / sin(t)^4 =>

sin(t)^(1/3) * cos(t)^(2/3) * cos(t) * dt / sin(t)^4 =>

sin(t)^(-11/3) * cos(t)^(5/3) * dt =>

cos(t)^(5/3) * dt / sin(t)^(11/3) =>

(cos(t) / sin(t))^(5/3) * dt / sin(t)^(6/3) =>

(cos(t) / sin(t))^(5/3) * csc(t)^2 * dt =>

cot(t)^(5/3)* csc(t)^2 * dt

cot(t) = m

dm = -csc(t)^2 * dt

-m^(5/3) * dm

Integrate

-m^(8/3) * (3/8) + C =>

(-3/8) * m^(8/3) + C =>

(-3/8) * cot(t)^(8/3) + C =>

(-3/8) * cos(t)^(8/3) / sin(t)^(8/3) + C =>

(-3/8) * (sqrt(1 - sin(t)^2))^(8/3) / sin(t)^(8/3) + C =>

(-3/8) * (1 - sin(t)^2)^(4/3) / sin(t)^(8/3) + C =>

(-3/8) * (1 - x^2)^(4/3) / x^(8/3) + C =>

(-3/8) * (1 - x^2)^(4/3) / (x^2)^(4/3) + C =>

(-3/8) * ((1 - x^2) / x^2)^(4/3) + C

From 1/3 to 1

(-3/8) * ((1 - (1)^2 / 1^2)^(4/3) + (3/8) * ((1 - (1/3)^2) / (1/3)^2)^(4/3) =>

(3/8) * (((8/9) / (1/9))^(4/3) - ((0/1)^(4/3)) =>

(3/8) * (8/1)^(4/3) =>

(3/8) * 2^4 =>

(3/8) * 16 =>

3 * 2 =>

6

∫ = ∫ (x−x³)^(1/3)] / x⁴dx, [x=⅓,1]

Looking at the form of Captain's answer I couldn't help but notice …

∫ = ∫ (x‾²−1)^(1/3) x‾³ dx = ∫ (x‾²−1)^(1/3) (-½)d(x‾²−1) dx = -½ * ¾ * (x‾²−1)^(4/3)