\(\frac 1{(x^2 + ax + a^2)(x^2+bx + b^2)} = \frac{Ax + B}{x^2 + ax+a^2}+\frac{Cx + D}{x^2 + bx + b^2}\)
\(\therefore 1 = (Ax + B) (x^2 + bx + b^2) + (Cx +D) (x^2 + ax + a^2)\)
Also,
\((A + C)x^3 + (Ab+B + aC + D)x^2 \\+(Ab^2 + Bb + Ca^2 + Da)x + (Bb^2 + Da^2) = 1\)
\(\therefore A + C =0\) .....(1)
\(Ab + B+ Ca +D = 0\) .....(2)
\(Ab^2 + Bb + Ca^2 + Da = 0\) .....(3)
\(Bb^2 + Da^2 = 1\) ......(4)
From (3),
\((Ab + B)b + (Ca+ D)a = 0\)
\(\therefore Ca + D = -\frac{b}a (Ab + B) \)
From (2),
\((Ab + B) - \frac ba (Ab + B) = 0\)
⇒ \((1 - \frac ba) (Ab + B) = 0\)
⇒ \(Ab + B = 0\)
⇒ \(B = -Ab\) \((\because a \ne b)\)
\(\therefore Ca + D = 0 \) (From (2))
⇒ \(D = -Ca\)
From (4),
\(-Ab^3 - Ca^3 = 1\)
\(Ab^3 + Ca^3 = -1\)
\(-Cb^3 + Ca^3 = - 1\) (From(1))
⇒ \(C(a^3 - b^3) = -1\)
\(\therefore C = \frac 1{b^3 - a^3} = \frac{-1}{a^3 - b^3}\)
\(\therefore A = -C = \frac 1{a^3 - b^3}\)
\(B = -Ab = \frac{-b}{a^3 - b^3}\)
\(D = -Ca = \frac a{a^3 - b^3}\)
\(\therefore \frac 1{(x^2 + ax + b^2)(x^2 + bx + b^2)} = \frac 1{a^3 - b^3} \frac{x-b}{x^2 + ax + b^2} - \frac 1{a^3 - b^3}\frac{x-a}{^2 + bx + b^2}\)
\(\therefore \int\limits_{-\infty}^{\infty} \frac{dx}{(x^2 + ax + a^2 )(x^2 + bx + b^2)} = \frac 1{2(a^3 - b^3)} \int\limits_{-\infty}^\infty \frac{2x-2b}{x^2+ ax+a^2 } - \frac1{2(a^3 - b^3)} \int \limits_{-\infty}^\infty \frac{2x-2a}{x^2 + bx + b^2}dx\)
\(= \frac1{2(a^3 - b^3)} \left[\log (x^2 + ax + a^2)- \log(x^2 + bx + b^2)\right]_{-\infty}^\infty \\- \frac{2b+a}{2(a^3 - b^3)} \int\limits_{-\infty}^\infty \cfrac{dx}{(x + \frac a2)^2 + \frac{3a^2}4}+\frac{2a +b}{2(a^3 - b^3)} \int\limits_{-\infty}^{\infty} \cfrac{dx}{(x + \frac b2)^2 + \frac{3b^2}4}\)
\(= \frac1{2(a^3 - b^3)}\left[\log\left(\cfrac{1 + \frac ax + \frac{a^2}{x^2}}{1 + \frac bx + \frac{b^2}{x^2}}\right)\right] _{-\infty}^\infty- \frac{2b +a}{2(a^3 - b^3)} \times \frac 1{\frac{\sqrt 3a}2} \left[ \tan^{-1} \left(\cfrac{x + \frac a2}{\frac{\sqrt3a}2}\right)\right]_{-\infty}^\infty\\ + \frac{2a + b}{2(a^3 - b^3)} \times \frac 1{\frac{\sqrt 3b}2} \left[ \tan^{-1} \left(\cfrac{x + \frac b2}{\frac{\sqrt3b}2}\right)\right]_{-\infty}^\infty \)
\(= \frac 1{2(a^3 - b^3)} [\log 1]_{-\infty}^\infty - \frac{2b +a}{\sqrt 3a (a^3 - b^3)}\times (\tan^{-1}\infty - \tan^{-1}(-\infty))\\+ \frac{2a + b}{\sqrt 3 b(a^3 - b^3)}(\tan^{-1}{\infty}-\tan^{-1}-\infty)\)
\(= 0 - \frac{2b +a}{\sqrt 3 a(a^3 - b^3)}\times \left(\frac \pi 2+ \frac \pi 2\right) + \frac{2a +b}{\sqrt 3b(a^3 -b^3)}\left(\frac \pi 2 + \frac \pi 2\right)\)
\(= \frac{-(2b +a)\pi}{\sqrt 3 a (a^3 - b^3)} + \frac{(2a + b)\pi}{\sqrt 3b(a^3 - b^3)}\)
\(= \frac{\pi (-2b^2 - ab + 2a^2 + ab)}{\sqrt 3 ab(a^3 - b^3)}\)
\(= \frac{2\pi(a - b)(a + b)}{\sqrt 3 ab (a -b)(a^2 +ab + b^2)}\)
\(= \frac{2\pi (a + b)}{\sqrt 3 ab (a^2 + ab+ b^2)}\)
Hence proved.
