$\int {\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} dx$

Let $I = \int {\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} dx$

Now, $\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}$

[let ${x^2} = t$ ]
$= \frac{t}{{\left( {t + {a^2}} \right)\left( {t + {b^2}} \right)}} = \frac{A}{{\left( {t + {a^2}} \right)}} + \frac{B}{{\left( {t + {b^2}} \right)}}$

$t = A\left( {t + {b^2}} \right) + B\left( {t + {a^2}} \right)$

By comparing the coefficients of $t$, we get
$A + B = 1$ …….(i)

${b^2}A + {a^2}B = 0$ ………(ii)

$\Rightarrow$ ${b^2}(1 - B) + {a^2}B = 0$

$\Rightarrow$ ${b^2} - {b^2}B + {a^2}B = 0$

$\Rightarrow$ ${b^2} + \left( {{a^2} - {b^2}} \right)B = 0$

$\Rightarrow$ $B = \frac{{ - {b^2}}}{{{a^2} - {b^2}}} = \frac{{{b^2}}}{{{b^2} - {a^2}}}$

From Eq. (i), $A + \frac{{{b^2}}}{{{b^2} - {a^2}}} = 1$

$\Rightarrow$ $A = \frac{{{b^2} - {a^2} - {b^2}}}{{{b^2} - {a^2}}} = \frac{{ - {a^2}}}{{{b^2} - {a^2}}}$

therefore, $I = \int {\frac{{ - {a^2}}}{{\left( {{b^2} - {a^2}} \right)\left( {{x^2} + {a^2}} \right)}}} dx + \int {\frac{{{b^2}}}{{{b^2} - {a^2}}}} \cdot \frac{1}{{{x^2} + {b^2}}}dx$

$= \frac{{ - {a^2}}}{{\left( {{b^2} - {a^2}} \right)}}\int {\frac{1}{{{x^2} + {a^2}}}} dx + \frac{{{b^2}}}{{{b^2} - {a^2}}}\int {\frac{1}{{{x^2} + {b^2}}}} dx$

$= \frac{{ - {a^2}}}{{{b^2} - {a^2}}} \cdot \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + \frac{{{b^2}}}{{{b^2} - {a^2}}} \cdot \frac{1}{b}{\tan ^{ - 1}}\frac{x}{b}$

$= \frac{1}{{{b^2} - {a^2}}}\left[ { - a\,ta{n^{ - 1}}\frac{x}{a} + b{{\tan }^{ - 1}}\frac{x}{b}} \right]$

$= \frac{1}{{{a^2} - {b^2}}}\left[ {{{{\mathop{\rm atan}\nolimits} }^{ - 1}}\frac{x}{a} - b{{\tan }^{ - 1}}\frac{x}{b}} \right]$