Explanation

(c)
Let $I = \int_{}^{}\frac{(x^{2} - 1)dx}{x^{3}\sqrt{2x^{4} - 2x^{2} + 1}}$
On dividing Nr and Dr by $x^{5}$, we get
$I = \int_{}^{}\frac{\left( \frac{1}{x^{3}} - \frac{1}{x^{5}} \right)dx}{\sqrt{2 - \frac{2}{x^{2}} + \frac{1}{x^{4}}}}$
Put $2 - \frac{2}{x^{2}} + \frac{1}{x^{4}} = t\ \ \ \Rightarrow \ \ \ \ \left( \frac{4}{x^{3}} - \frac{4}{x^{5}} \right)dx = dt$
$\ \therefore I = \frac{1}{4}\ \int_{}^{}\frac{dt}{\sqrt{t}} = \frac{1}{2}\sqrt{t} + c = \frac{1}{2}\sqrt{2 - \frac{2}{x^{2}} + \frac{1}{x^{4}}} + c$