Explanation

(a)
Let $I = \int_{}^{}\frac{dx}{x^{2}\left( x^{4} + 1 \right)^{3/4}} = \int_{}^{}\frac{dx}{x^{2}.x^{3}\left( 1 + \frac{1}{x^{4}} \right)^{3/4}}$
Put $1 + x^{- 4} = t\ \ \ \ \Rightarrow \ \ \frac{- 4}{x^{5}}\ dx = dt$
$\therefore\ \ \ I = - \frac{1}{4}\int_{}^{}\frac{dt}{t^{3/4}} = - \frac{1}{4}.\ \frac{t^{1/4}}{1/4} + c$
$= - \left( 1 + \frac{1}{x^{4}} \right)^{1/4} + c$
$= \frac{- \left( x^{4} + 1 \right)^{1/4}}{x} + c$