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

1 Answer
Oct 18, 2017

# 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#