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.
