Question
$$\int\limits_0^\infty {\frac{{dx}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} {\text{ is ?}}$$
A.
$$\frac{{\pi ab}}{{a + b}}$$
B.
$$\frac{\pi }{{2\left( {a + b} \right)}}$$
C.
$$\frac{\pi }{{2ab\left( {a + b} \right)}}$$
D.
$$\frac{{\pi \left( {a + b} \right)}}{{2ab}}$$
Answer :
$$\frac{\pi }{{2ab\left( {a + b} \right)}}$$
Solution :
$$\eqalign{
& \int\limits_0^\infty {\frac{{dx}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} \cr
& {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\int\limits_0^\infty {\frac{{\left( {{x^2} + {b^2}} \right) - \left( {{x^2} + {a^2}} \right)}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} \cr
& {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\int\limits_0^\infty {\left[ {\frac{1}{{{x^2} + {a^2}}} - \frac{1}{{{x^2} + {b^2}}}} \right]dx} \cr
& {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\left[ {\frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a} - \frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right]_0^\infty \cr
& {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\left[ {\frac{\pi }{{2a}} - \frac{\pi }{{2b}}} \right] \cr
& {\text{ = }}\frac{\pi }{{2ab\left( {a + b} \right)}} \cr} $$