Improper Integral Involving A Probable Partial Fraction Conversion

Question

I was having trouble evaluating the following integral:$$\int^{+\infty}_{-\infty} \frac{dx}{(x^2\pm ax+a^2)(x^2\pm bx+b^2)}$$ I have tried approaches that involve converting into partial fractions using thumb rule and without using it.I also tried the online integral calculator. No luck whatsoever. The answer according to the book is: $$\frac{2\pi}{\sqrt{3}}.\frac{a+b}{ab(a^2+ab+b^2)}$$

Answer 1

We have $$\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} = \frac{cx+d}{x^2+ax+a^2} + \frac{ex+f}{x^2+bx+b^2}.$$ Solving for $c,d,e,f$ one should find \begin{align*} \frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} &= \frac{1}{(a-b)(a^2+ab+b^2)}\left( \frac{x-b}{x^2+ax+a^2} + \textrm{something similar}\right). \end{align*} Now use the hint in the answer by @Dr. Sonnhard Graubner.

Addendum:

Letting $x=t-a/2$ we find $$\frac{x-b}{x^2+ax+a^2} = \frac{t-(b+a/2)}{t^2+3a^2/4}.$$ The integral \begin{align*} \int_{-\infty}^\infty \frac{dt}{t^2+3a^2/4} \tag{1} \end{align*} is a standard integral. The integral \begin{align*} \int_{-\infty}^\infty \frac{t dt}{t^2+3a^2/4} \tag{2} \end{align*} is divergent. Letting $x=t-b/2$ in the "something similar" above we find terms like (1) and (2) above. Call them (3) and (4), respectively. Combining (2) and (4) we find we must integrate $$\frac{1}{t^2+3a^2/4}-\frac{1}{t^2+3b^2/4} = (\mathrm{const})\frac{t}{(t^2+3a^2/4)(t^2+3b^2/4)}.$$ This function is integrable on $(-\infty,\infty)$ and odd and therefore integrates to zero.

Answer 2

HINT: use that $$x^2+ax+a^2=\left(x+\frac{a}{2}\right)^2+\frac{3}{4}a^2$$ and $$x^2+bx+b^2=\left(x+\frac{b}{2}\right)^2+\frac{3}{4}b^2$$