Find the most general antiderivative of the function.(a) $f ( x ) = 12 x ^ { 2 } + 6 x - 5$ (b) $f ( x ) = \left( 1 - \frac { 1 } { \sqrt { x } } \right) \left( 1 + \frac { 1 } { \sqrt { x } } \right)$ (c) $f ( x ) = \frac { 1 } { x ^ { 2 } } - \frac { 2 } { x ^ { 3 } }$ (d) $f ( t ) = \frac { t ^ { 3 } + t ^ { 2 } } { \sqrt { t } }$

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The Answer of Find the most general antiderivative of the function.(a) $f ( x ) = 12 x ^ { 2 } + 6 x - 5$ (b) $f ( x ) = \left( 1 - \frac { 1 } { \sqrt { x } } \right) \left( 1 + \frac { 1 } { \sqrt { x } } \right)$ (c) $f ( x ) = \frac { 1 } { x ^ { 2 } } - \frac { 2 } { x ^ { 3 } }$ (d) $f ( t ) = \frac { t ^ { 3 } + t ^ { 2 } } { \sqrt { t } }$

Find the Most General Antiderivative of the Function f(x)=12x2+6x−5f ( x ) = 12 x ^ { 2 } + 6 x - 5f(x)=12x2+6x−5

Find the Most General Antiderivative of the Function $f ( x ) = 12 x ^ { 2 } + 6 x - 5$