To get the partial fraction, we need to factorize the denominator, which is #x^4-x^2-1#. To do so, first observe that there are no terms with degree 1 or 3. Hence, we can make a substitution of #y=x^2#. So,
#x^4-x^2-1=y^2-y-1#,
which becomes a quadratic expression. Using either the quadratic formula or by completing the square, the roots to the equation #y^2-y-1=0# are #y = frac{1+sqrt{5}}{2}# and #y = frac{1-sqrt{5}}{2}#.
Therefore,
#x^4-x^2-1=y^2-y-1#
#=(y-frac{1+sqrt{5}}{2})(y-frac{1-sqrt{5}}{2})#
#=(x^2-frac{1+sqrt{5}}{2})(x^2-frac{1-sqrt{5}}{2})#
#=(x+sqrt{frac{1+sqrt{5}}{2}})(x-sqrt{frac{1+sqrt{5}}{2}})(x^2+frac{sqrt{5}-1}{2})#.
Now that the denominator has been completely factorized, we can proceed with the partial fractions. We first note that the numerator is of a smaller degree than the denominator. We can thus write
#frac{x^3}{x^4-x^2-1}-=frac{A}{x+sqrt{frac{1+sqrt{5}}{2}}}#
#+frac{B}{x-sqrt{frac{1+sqrt{5}}{2}}}+frac{Cx+D}{x^2+frac{sqrt{5}-1}{2}}#,
where #A#, #B#, #C# and #D# are real constants to be determined.
By multiplying both sides with #(x^4-x^2-1)#, we get
#x^3-=A(x-sqrt{frac{1+sqrt{5}}{2}})(x^2+frac{sqrt{5}-1}{2})#
#+B(x+sqrt{frac{1+sqrt{5}}{2}})(x^2+frac{sqrt{5}-1}{2})#
#+(Cx+D)(x^2-frac{1+sqrt{5}}{2})#.
To find #A#, substitute a value of #x# that will cancel out the terms with #B#, #C# and #D#. In this case, substitute #x=-sqrt{frac{1+sqrt{5}}{2}}#.
#(-sqrt{frac{1+sqrt{5}}{2}})^3=A((-sqrt{frac{1+sqrt{5}}{2}})#
#-sqrt{frac{1+sqrt{5}}{2}})((-sqrt{frac{1+sqrt{5}}{2}})^2+frac{sqrt{5}-1}{2})#
We get #A=frac{5+sqrt{5}}{20}#.
Similarly, to find #B#, substitute a value of #x# that will cancel out the terms with #A#, #C# and #D#. In this case, substitute #x=sqrt{frac{1+sqrt{5}}{2}}#.
#(sqrt{frac{1+sqrt{5}}{2}})^3=B((sqrt{frac{1+sqrt{5}}{2}})#
#+sqrt{frac{1+sqrt{5}}{2}})((sqrt{frac{1+sqrt{5}}{2}})^2+frac{sqrt{5}-1}{2})#
We get #B=frac{5+sqrt{5}}{20}#.
Now to find #C# and #D#, the approach is slightly different. To find #D#, we need to find a real number #x# to substitute such that the terms with #A#, #B# and #C# will disappear. However, there are no such #x#. But not to worry, since we have already found #A# and #B#, a value of #x# that will eliminate the terms with #C# only will suffice. In this case, substitute #x=0# .
#(0)^3=frac{5+sqrt{5}}{20}((0)-sqrt{frac{1+sqrt{5}}{2}})((0)^2+frac{sqrt{5}-1}{2})#
#+frac{5+sqrt{5}}{20}((0)+sqrt{frac{1+sqrt{5}}{2}})((0)^2+frac{sqrt{5}-1}{2})#
#+(C(0)+D)((0)^2-frac{1+sqrt{5}}{2})#
We get #D=0#.
To find #C#, we can compare the coefficients of the #x^3# term.
#1=A+B+C#
We get #C=frac{5-sqrt{5}}{10}#.
Hence,
#frac{x^3}{x^4-x^2-1}-=frac{frac{5+sqrt{5}}{20}}{x+sqrt{frac{1+sqrt{5}}{2}}}#
#+frac{frac{5+sqrt{5}}{20}}{x-sqrt{frac{1+sqrt{5}}{2}}}+frac{(frac{5-sqrt{5}}{10})x}{x^2+frac{sqrt{5}-1}{2}}#.
Now, we proceed with the integration.
#intfrac{x^3}{x^4-x^2-1}dx=intfrac{frac{5+sqrt{5}}{20}}{x+sqrt{frac{1+sqrt{5}}{2}}}dx#
#+intfrac{frac{5+sqrt{5}}{20}}{x-sqrt{frac{1+sqrt{5}}{2}}}dx+intfrac{(frac{5-sqrt{5}}{10})x}{x^2+frac{sqrt{5}-1}{2}}dx#
#=frac{5+sqrt{5}}{20}ln|x+sqrt{frac{1+sqrt{5}}{2}}|+frac{5+sqrt{5}}{20}ln|x-sqrt{frac{1+sqrt{5}}{2}}|#
#+intfrac{(frac{5-sqrt{5}}{10})x}{x^2+frac{sqrt{5}-1}{2}}dx#
#=frac{5+sqrt{5}}{20}ln|x^2-frac{1+sqrt{5}}{2}|+intfrac{(frac{5-sqrt{5}}{10})x}{x^2+frac{sqrt{5}-1}{2}}dx#
#=frac{5+sqrt{5}}{20}ln|x^2-frac{1+sqrt{5}}{2}|+(frac{5-sqrt{5}}{20})intfrac{2x}{x^2+frac{sqrt{5}-1}{2}}dx#
#=frac{5+sqrt{5}}{20}ln|x^2-frac{1+sqrt{5}}{2}|+frac{5-sqrt{5}}{20}ln(x^2+frac{sqrt{5}-1}{2})#
#+c#, where #c# is the constant of integration.
