Verified answer

∫ v/(6v³ - 1) dv

= (1/6) ∙ ∫ v/(v³ - (1/6)) dv

You can solve by expanding to partial fractions.

First factorize denominator.

one root is at

v = (∛(1/6)

To improve readability i use symbol a=∛(1/6) in the following

By long division i find

(v³ - a³) / (v - a) = v² + a∙v + a²

<=>

(v³ - a³) = (v - a)∙(v² + a∙v + a²)

Hence you can decompose as follows

v/(v³ - a³) = A/(v - a) + (B∙v + C)/(v² + a∙v + a²)

<=>

v = A∙(v² + a∙v + a²) + B∙v∙(v - a) + C∙(v - a)

<=>

v = v²(A + B) + v∙(a∙A - a∙B + C) + 1∙(a²∙A - a∙C)

Compare coefficients of same powers of v and find

A + B = 0

a∙A - a∙B + C = 1

a²∙A - a∙C = 0

Solution to this equation system is

A = 1/(3∙a)

B = -1/(3∙a)

C = 1/3

=>

(1/6) ∙ ∫ v/(v³ - (1/6)) dv

= a³ ∙ ∫ v/(v³ - a³) dv

= a³ ∙ ∫ [1/(3∙a)]/(v - a) - ([1/(3∙a)]∙v - [1/3])/(v² + a∙v + a²) dv

= (a²/3) ∙ ∫ 1/(v - a) - (v - a)/(v² + a∙v + a²) dv

= (a²/3) ∙ [ ∫ 1/(v - a) dv - ∫ (v - a)/(v² + a∙v + a²) dv ]

= (a²/3) ∙ [ ln(v - a) - ∫ (v - a)/(v² + a∙v + a²) dv ]

To solve the remaining integral use quadratic supplement to bring you denominator to the form z² + 1. By appropriate substitution you get two integrals of the from:

∫ z/(z² + 1) dz = (1/2)∙ln(z² +1)

and

∫ 1/(z² + 1) dz = tan⁻¹(z)

∫ (v - a)/(v² + a∙v + a²) dv

= ∫ (v - a)/( v² + a∙v + (1/4)∙a² + (3/4)∙a²) dv

= ∫ (v - a)/( (v + (1/2)∙a)² + (3/4)∙a² ) dv

= (4/(3∙a²) ∙ ∫ (v - a)/( [(2/√3)∙(v + (1/2)∙a)/a]² + 1 ) dv

= (4/(3∙a²) ∙ ∫ (v - a)/( [(2∙(v/a) + 1)/√3]² + 1 ) dv

let

z = [(2∙(v/a) + 1)/√3]

<=>

v = (z∙√3 - 1)∙(a/2)

=>

dv = (a∙√3)/2 dz

(4/(3∙a²) ∙ ∫ (v - a) / ( [(2∙(v/a) + 1)/√3]² + 1 ) dv

= (2/(a∙√3) ∙ ∫ ((z∙√3 - 1)∙(a/2) - a ) / ( z² + 1 ) dz

= (2/(a∙√3) ∙ ∫ ( z∙√3∙(a/2) - (3/2)∙a ) / ( z² + 1 ) dz

= ∫ z/(z² + ) - √3/(z² + 1) dz

= ∫ z/(z² + ) - √3/(z² + 1) dz

= (1/2)∙ln(z² +1) - √3∙tan⁻¹(z)

substitute back

z = [(2∙(v/a) + 1)/√3]

z² + 1 = v² + a∙v + a²

(1/2)∙ln(z² +1) - √3∙tan⁻¹(z)

= (1/2)∙ln(v² + a∙v + a²) - √3∙tan⁻¹((2∙(v/a) + 1)/√3)

Summary:

∫ v/(6v³ - 1) dv

= a³ ∙ ∫ v/(v³ - a³) dv

= (a²/3) ∙ { ln(v - a) - (1/2)∙ln(v² + a∙v + a²) + √3∙tan⁻¹((2∙(v/a) + 1)/√3) }

= { ln(v - ∛(1/6)) - (1/2)∙ln(v² + ∛(1/6)∙v + ∛(1/36)) + √3∙tan⁻¹((2∙∛6∙v + 1)/√3) } / (3∙∛6)