1
Given that
\( \dfrac{d y}{d x} = 3x^2 - \dfrac{2 - 3x}{x^3}, \quad x \neq 0 \)
and \(y = 5\) when \(x = 1\), find \(y\) in terms of \(x\).
A
\(y = \dfrac{1}{3} x^3 + x^{-2} - 3 x^{-1} + 6 \dfrac{2}{3}\)
B
\(y = x^3 + \dfrac{1}{2} x^{-2} - 3 x^{-1} + 6 \dfrac{1}{2}\)
C
\(y = x^3 + x^{-2} - 3 x^{-1} + 6\)
D
\(y = x^3 + x^{-2} - x^{-1} + 4\)
E
\(y = 3 x^3 + x^{-2} - x^{-1} + 2\)