Question

Evaluate the integral `int \ x/(a^3-x^3) \ dx`?

Answer 1

`int \ x/(a^3-x^3) \ dx = 1/(6a) ln|x^2+ax+a^2| -sqrt(3)/(3a) \ arctan( (2x+a)/(sqrt(3)a)) - 1/(3a) \ ln|x-a| + C`

Explanation:

We seek:

`I = int \ x/(a^3-x^3) \ dx`

We can factorize the integrand, and then decompose into partial fractions:

`x/(a^3-x^3) = x/((a-x)(x^2+ax+a^2))`

`" " = A/(a-x) + (Bx+C)/(x^2+ax+a^2)`

`" " = ( A(x^2+ax+a^2) + (Bx+C)(a-x) ) / ((a-x)(x^2+ax+a^2))`

Leading to the identity:

`x -= A(x^2+ax+a^2) + (Bx+C)(a-x)`

Where `A,B,C` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = a => a = A(a^2+a^2+a^2) => A = 1/(3a)`
`Coeff(x^2): 0 = A-B => B=1/(3a)`
`Coeff(x^0): 0 = Aa^2+aC => C = -1/3`

So, we have:

`x/(a^3-x^3) -= ( 1/(3a) )/(a-x) + ((1/3a)x-1/3)/(x^2+ax+a^2)`
`" " = 1/(3a) { 1/(a-x) + (x-a)/(x^2+ax+a^2) }`
`" " = 1/(3a) { (x-a)/(x^2+ax+a^2) - 1/(x-a) }`

So we can write the integral as:

`I = 1/(3a) int \ (x-a)/(x^2+ax+a^2) - 1/(x-a) \ dx`
`\ \ = 1/(3a) { int \ (x-a)/(x^2+ax+a^2) \ dx - int \ 1/(x-a) \ dx }`
`\ \ = 1/(3a) int \ (x-a)/(x^2+ax+a^2) \ dx - 1/(3a) \ ln|x-a| + C`
`\ \ = 1/(3a) \ I_2 - 1/(3a) \ ln|x-a| + C`

Now, let us consider the remaining integral, `I_2`, which requires a little manipulation, as follows:

`I_2 = int \ (x-a)/(x^2+ax+a^2) \ dx`
`\ \ \ = 1/2 \ int \ (2(x-a))/(x^2+ax+a^2) \ dx`
`\ \ \ = 1/2 \ int \ (2x-2a)/(x^2+ax+a^2) \ dx`
`\ \ \ = 1/2 \ int \ (2x+a-3a)/(x^2+ax+a^2) \ dx`
`\ \ \ = 1/2 \ int \ (2x+a)/(x^2+ax+a^2) -(3a)/(x^2+ax+a^2)\ dx`
`\ \ \ = 1/2 \ int \ (2x+a)/(x^2+ax+a^2) -(3a)/2 \ int \ (1)/(x^2+ax+a^2) \ dx`
`\ \ \ = 1/2 \ ln|x^2+ax+a^2| -(3a)/2 I_3`

Next, we consider, `I_3`, where

`I_3 = int \ (1)/(x^2+ax+a^2) \ dx`
`\ \ \ = int \ (1)/((x+a/2)^2 - (a/2)^2+a^2) \ dx`
`\ \ \ = int \ (1)/((x+a/2)^2 - a^2/4+a^2) \ dx`
`\ \ \ = int \ (1)/((x+a/2)^2 +(3a^2)/4) \ dx`
`\ \ \ = int \ (1)/( (3a^2)/4( (4/(3a^2))(x+a/2)^2 + 1) )\ dx`
`\ \ \ = 4/(3a^2) \ int \ (1)/( (4/(3a^2))(x+a/2)^2 + 1) \ dx`
`\ \ \ = 4/(3a^2) \ int \ (1)/( ((2/(sqrt(3)a))(x+a/2))^2 + 1) \ dx`
`\ \ \ = 4/(3a^2) \ int \ (1)/( (2/(sqrt(3)a)x +1/sqrt(3))^2 + 1) \ dx`

Now we can perform a substitution, Let

`u = 2/(sqrt(3)a)x +1/sqrt(3) => (du)/dx = 2/(sqrt(3)a)`

Then substituting into `I_3` we get:

`I_3 = 4/(3a^2) \ int \ (1)/( u^2 + 1) \ ((sqrt(3)a)/2) \ du`
`\ \ \ = (2sqrt(3))/(3a) \ int \ (1)/( u^2 + 1) \ du`
`\ \ \ = (2sqrt(3))/(3a) \ arctan(u)`
`\ \ \ = (2sqrt(3))/(3a) \ arctan(2/(sqrt(3)a)x +1/sqrt(3))`
`\ \ \ = (2sqrt(3))/(3a) \ arctan( (2x+a)/(sqrt(3)a))`

Combining this result with `I_2` we get:

`I_2 = 1/2 \ ln|x^2+ax+a^2| -(3a)/2 I_3`
`\ \ \ = 1/2 \ ln|x^2+ax+a^2| -(3a)/2 (2sqrt(3))/(3a) \ arctan( (2x+a)/(sqrt(3)a))`
`\ \ \ = 1/2 \ ln|x^2+ax+a^2| -sqrt(3) \ arctan( (2x+a)/(sqrt(3)a))`

And finally combining this result with `I` we get

`I = 1/(3a) \ I_2 - 1/(3a) \ ln|x-a| + C`
`\ \ = 1/(3a) {1/2 \ ln|x^2+ax+a^2| -sqrt(3) \ arctan( (2x+a)/(sqrt(3)a))} - 1/(3a) \ ln|x-a| + C`

`\ \ = 1/(6a) ln|x^2+ax+a^2| -sqrt(3)/(3a) \ arctan( (2x+a)/(sqrt(3)a)) - 1/(3a) \ ln|x-a| + C`