Verified answer

Notice that it is the difference of two cubes and factor it using:

a³ - b³ = ( a - b )·( a² + a·b + b² )

————

∫1/(x³ - a³) dx

= ∫1/[ ( x - a )·( x² + a·x + a² ) ] dx

————

Use partial fractions:

1/[ ( x - a )·( x² + a·x + a² ) ] = A/( x - a ) + B/( x² + a·x + a² )

1 = A·( x² + a·x + a² ) + B·( x - a )

Choose a convenient value for x to cancel either A or B. Choose x=a:

1 = A·( (a)² + a·(a) + a² ) + B·( (a) - a )

1 = A·( 3·a² )

1/( 3·a² ) = A

Use this value to solve for B:

1 = 1/( 3·a² ) · ( x² + a·x + a² ) + B·( x - a )

-B·( x - a ) = 1/( 3·a² ) · ( x² + a·x + a² ) - 1

-B·3·a²·( x - a ) = x² + a·x - 2·a²

-B·3·a²·( x - a ) = ( x - a )·( x + 2·a )

B = -( x + 2·a ) / ( 3·a² )

————

∫ [ A/( x - a ) + B/( x² + a·x + a² ) ] dx

→ ∫ [ 1/( 3·a² ) /( x - a ) - ( x + 2·a ) / ( 3·a² ) / ( x² + a·x + a² ) ] dx

= 1/( 3·a² )·∫1/( x - a )dx - 1/( 3·a² )·∫( x + 2·a )/( x² + a·x + a² )dx

This are two integrals now.

(I'll leave out the factored-out constants while evaluating.)

————

Integral one:

∫1/( x - a ) dx

Let u = x - a

Then du = dx

→ ∫1/u du

= ln|u| + C,

————

Integral two:

∫( x + 2·a )/( x² + a·x + a² )]dx

= ½·∫( 2·x + a + 3·a )/( x² + a·x + a² )]dx

= ½·∫( 2·x + a )/( x² + a·x + a² )]dx + ½·∫( 3·a )/( x² + a·x + a² )]dx

Let v = x² + a·x + a²

Then dv = (2·x + a)dx

→ ½·∫dv/v + ½·∫( 3·a )/( x² + a·x + a² )]dx

= ½·ln| v | + 3·a/2·∫1/( x² + a·x + a² )]dx

————

Last part is sadly the hardest.

You want to make a trigonometric substitution.

Complete the square in the denominator:

∫1/( x² + a·x + a² ) dx

= ∫1/( (x² + a·x + (a/2)² - (a/2)² + a² ) dx

= ∫1/[(x + a/2)² + 3·a²/4 ] dx

Let q = x + a/2

Then dq = dx

→ ∫1/( q² + 3·a²/4 ) dq

Get 3·a²/4 to be just 1. Multiply by 1/1:

= [4/(3·a²)]/[ 4/(3·a²) ] · ∫1/( q² + 3·a²/4 ) dq

= [ 4/(3·a²) ] · ∫1/{ [4/(3·a²)]·q² + 1 ] dq

Bring the the coefficient of q² into the square by getting its square root:

= [ 4/(3·a²) ] · ∫1/{ [2/(√(3)·a)·q]² + 1 ] dq

Let tan(r) = 2/[√(3)·a]·q

Then r = arctan[ 2/[√(3)·a]·q ]

And dr = 1 / { [ 2/[√(3)·a]·q ]² + 1 } · 2/[√(3)·a] · dq

And dq = { [ 2/[√(3)·a]·q ]² + 1 } · √(3)·a/2 · dr

Replace dq and reduce:

→ [ 4/(3·a²) ] · ∫1/{ [2/(√(3))·a·q]² + 1 ]·{ [ 2/[√(3)·a]·q ]² + 1 } · √(3)·a/2·dr

= [ 4/(3·a²) ] · ∫ √(3)·a/2 dr

= 2/[ √(3)·a ] · r + C,,

————

Put these integrals all back together with the left-out constants:

1/( 3·a² )·∫1/( x - a )dx - 1/( 3·a² )·∫( x + 2·a )/( x² + a·x + a² )dx

= 1/( 3·a² )·{ ln|u| } - 1/( 3·a² )·{ ½·ln| v | + 3·a/2·2/[ √(3)·a ]·r } + C

Reverse the substitutions for u, v, and r (and then for q inside r):

= 1/( 3·a² )·{ ln|x - a| } - 1/( 3·a² )·{ ½·ln|x² + a·x + a²| + 3·a/2·2/[ √(3)·a ]·arctan[ 2/[√(3)·a]·q ] } + C

= 1/( 3·a² )·{ ln|x - a| } - 1/( 3·a² )·{ ½·ln|x² + a·x + a²| + 3·a/2·2/[ √(3)·a ]·arctan[ 2/[√(3)·a]·(x + a/2) ] } + C

Simplify:

= 1/(3·a²)·{ ln|x - a| - ½·ln|x² + a·x + a²| - √(3)·arctan[(2·x/a + 1)/√(3)] } + C

I am not sure,

but, let me give a try!

∫(1/x³ - a³)dx = (1/(3x^2)) x log(x^3 - a^3)

(or)

∫(1/x³ - a³)dx = log(x^3 - a^3) / (3x^2))

Both answers here are same Only!

Hello

write (x³-a³)=(x-a)(x²+ax+a²)

Then find A,B and C such that

1/(x³-a³) = A/(x-a) + (Bx+C)/(x²+ax+a²)

Integration is immediate as you can see after Gravitate's work;-)

no thanks, it is totally horrible