Simplifying quadratic and higher degree expressions in partial fractions method

Question

While integrating several functions containing algebraic terms many times we use the partial fractions method, for example consider the integrand $$\frac{1}{x(x^4-1)} = \frac{A}{x} + \frac{B}{x-1}+ \frac{C}{x+1}+ \frac{Dx+E}{x^2+1}$$ Now, either we can re-multiply terms on RHS and compare coefficients or for linear terms we can apply a short 'trick' like for calculating $A$,

put $x=0$ in LHS removing the term making it undefined i.e. new LHS is $\frac{1}{x^4-1}$; $x=0$ gives $A=-1$ similarly $B=\frac{1}{4}, \ C=\frac{a1}{4}$ , hence I wonder if something similar can be done to get $D$ and $E$ .

NOTE: If it is restricted to be in the real domain, then I can't factorize $x^2+1=(x+i)(x-i)$ so, is there any short method to find constants like $D$ and $E$ for the quadratic expression and can we generalize a similar method for higher degree expressions like cubic etc. all help is greatly appreciated.

Answer 1

You had: $$1/x(x^4-1) = A/x + B/(x-1) + C/(x+1) + (Dx+E)/(x^2+1)$$ Multiply both sides of this by $x(x^4-1)$ to get: $$1 = A(x^4-1) + Bx(1 + x + x^2 + x^3) + Cx(x^2+1)(x-1) + (Dx + E)x(x^2-1)$$

Now plug in $x = i$, $$\implies 1 = 2D - 2Ei$$ And then plug in $x = -i$, $$\implies 1 = 2D + 2Ei$$

So you get $D = 1/2$ and $E = 0$

Answer 2

An easier method to evaluate thr integral is:$$\int \frac{1}{x(x^4-1)}dx=\int \frac{xdx}{x^2((x^2)^2-1)}$$ use the substitution:$u=x^2$ $$=\frac{1}{2}\int \frac{du}{u(u^2-1)}$$ I assume You can contionue form here