$\frac{{{{\rm{x}}^2}}}{{{{\rm{x}}^4} - 2{{\rm{x}}^2} - 15}}$
Or, $\frac{{{{\rm{x}}^2}}}{{{{\rm{x}}^4} - 2{{\rm{x}}^2} -
15}} = \frac{{\rm{y}}}{{{{\rm{y}}^2} - 2{\rm{y}} - 15}} =
\frac{{\rm{y}}}{{\left( {{\rm{y}} - 5} \right)\left( {{\rm{y}} + 3} \right)}}$
where y = x2
= $\frac{{\rm{A}}}{{{\rm{y}} - 5}} +
\frac{{\rm{B}}}{{{\rm{y}} + 3}}$.
or, y = A(y + 3) + B(y – 5).
Put y = 5,
5 = A.8
So, A = $\frac{5}{8}$.
Put y = -3,
Or, -3 = B(-8)
So, B = $\frac{3}{8}$.
So, $\frac{{\rm{y}}}{{{{\rm{y}}^2} - 2{\rm{y}} - 15}}$ =
$\frac{5}{8}.\frac{1}{{{\rm{y}} - 5}}$ + $\frac{3}{8}.\frac{1}{{4 + 3}}$.
i.e. $\frac{{{{\rm{x}}^2}}}{{{{\rm{x}}^4} - 2{{\rm{x}}^2} -
15}}$ = $\frac{5}{8}.\frac{1}{{{{\rm{x}}^2} - 5}} +
\frac{3}{8}.\frac{1}{{{{\rm{x}}^2} + 3}}$.
Or, $\mathop \smallint \nolimits^
\frac{{{{\rm{x}}^2}}}{{{{\rm{x}}^4} - 2{{\rm{x}}^2} - 15}}$.dx = $\mathop
\smallint \nolimits^ \left\{ {\frac{5}{8}.\frac{1}{{{{\rm{x}}^2} - 5}} +
\frac{3}{8}.\frac{1}{{{{\rm{x}}^2} + 3}}} \right\}$.dx
= $\frac{5}{8}$$\mathop \smallint \nolimits^
\frac{1}{{2\sqrt 5 }}\log \frac{{{\rm{x}} - \sqrt 5 }}{{{\rm{x}} + \sqrt 5 }} +
\frac{3}{8}.\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{{\rm{x}}}{{\sqrt 3 }}$ + c.
= $\frac{{\sqrt 5 }}{{16}}$.log $\frac{{{\rm{x}} - \sqrt 5 }}{{{\rm{x}} + \sqrt 5 }}$ + $\frac{{\sqrt 3 }}{8}$tan-1$\frac{{\rm{x}}}{{\sqrt 3 }}$ + c.