Explanation

(c)
Primitive function means indefinite integral.
$\therefore$ Primitive function of $f(x)$
$I = \int_{}^{}{\frac{\sqrt{(a^{2} - x^{2})}}{x^{4}}dx}$ (say)
$= \int_{}^{}{\frac{x\sqrt{\frac{a^{2}}{x^{2}} - 1}}{x \bullet x^{3}}dx}$
$= \int_{}^{}{\frac{1}{x^{3}} \bullet \sqrt{\left( \frac{a^{2}}{x^{2}} - 1 \right)}dx}$
Put $\frac{a^{2}}{x^{2}} - 1 = t^{2}$
$\Rightarrow \ - \frac{2a^{2}}{x^{3}}dx = 2t\ dt$
$\Rightarrow \frac{1}{x^{3}}dx = - \frac{1}{a^{2}}t\ dt$
Then, $I = - \frac{1}{a^{2}}\int_{}^{}{t^{2}dt}$
$= - \frac{1}{3a^{2}}t^{3} + c$
$= - \frac{1}{3a^{2}}\left( \frac{a^{2}}{x^{2}} - 1 \right)^{3/2} + c\$
$= - \frac{\left( a^{2} - x^{2} \right)^{3/2}}{3a^{2}x^{3}} + c$