Evaluate:: ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}dx$

$($i$)$ What is the correct trigonometric substitution to find an antiderivative:

$x=31sec()$

$x=961tan()$

$x=31sin()$

$x=961sec()$

$x=31tan()$

$($ii$)$ Which one is the correct statement after substituing the above trigonometric function:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}dx=\frac{1}{{31}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}se{c}^2()tan()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}dx=\frac{1}{{31}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^2()cos()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}dx=\frac{1}{31}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^2()cos()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}dx=31{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^2()cos()d$

$($iii$)$ What is the most general antiderivative of $\frac{\sqrt[\phantom{2}]{x^2-961}}{x^4}$