$\int {\frac{{{x^2}}}{{{x^4} - {x^2} - 12}}} dx$

Let $I = \int {\frac{{{x^2}}}{{{x^4} - {x^2} - 12}}} dx$

$= \int {\frac{{{x^2}}}{{{x^4} - 4{x^2} + 3{x^2} - 12}}} dx$

$= \int {\frac{{{x^2}dx}}{{{x^2}\left( {{x^2} - 4} \right) + 3\left( {{x^2} - 4} \right)}}}$

$= \int {\frac{{{x^2}dx}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 3} \right)}}}$

Now, $\frac{{{x^2}}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 3} \right)}}$

[let us assume : ${x^2} = t$]
$\Rightarrow$ $\frac{t}{{(t - 4)(t + 3)}} = \frac{A}{{t - 4}} + \frac{B}{{t + 3}}$

$\Rightarrow$ $t = A(t + 3) + B(t - 4)$

By comparing the coefficients of $t$ on both sides, we get
$A + B = 1$
………..(i)
$\Rightarrow$ $3A - 4B = 0$
………..(ii)

$\Rightarrow$ $3(1 - B) - 4B = 0$

$\Rightarrow$ $3 - 3B - 4B = 0$

$\Rightarrow$ $7B = 3$

$\Rightarrow$ $B = \frac{3}{7}$

If $B = \frac{3}{7},$ then $A + \frac{3}{7} = 1$

$\Rightarrow$ $A = 1 - \frac{3}{7} = \frac{4}{7}$

$\frac{{{x^2}}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 3} \right)}} = \frac{4}{{7\left( {{x^2} - 4} \right)}} + \frac{3}{{7\left( {{x^2} + 3} \right)}}$

therefore,$I = \frac{4}{7}\int {\frac{1}{{{x^2} - {{(2)}^2}}}} dx + \frac{3}{7}\int {\frac{1}{{{x^2} + {{(\sqrt 3 )}^2}}}} dx$

$= \frac{4}{7} \cdot \frac{1}{{2 \cdot 2}}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{3}{7} \cdot \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} + C$

$= \frac{1}{7}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{{\sqrt 3 }}{7}{\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} + C$