Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\ln \left( {1 + {{\left( {\ln 2} \right)}^4} + {{\rm{e}}^2}} \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = 4{x^2}\ln \left( {1 + {y^4} + {z^2}} \right)$ and the path:
${\bf{r}}\left( t \right) = \left( {{t^3},\ln \left( {1 + {t^2}} \right),{{\rm{e}}^t}} \right)$ for $0 \le t \le 1$
Evaluate:
$f\left( {{\bf{r}}\left( t \right)} \right) = 4{t^6}\ln \left( {1 + {{\left[ {\ln \left( {1 + {t^2}} \right)} \right]}^4} + {{\rm{e}}^{2t}}} \right)$
Since there is a potential function $f$ such that ${\bf{F}} = \nabla f$, we call ${\bf{F}}$ conservative. Thus, the line integral of ${\bf{F}}$ along the path ${\bf{r}}\left( t \right) = \left( {{t^3},\ln \left( {1 + {t^2}} \right),{{\rm{e}}^t}} \right)$ for $0 \le t \le 1$, depends only on the endpoints, that is
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{\bf{r}}\left( 1 \right)} \right) - f\left( {{\bf{r}}\left( 0 \right)} \right)$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\ln \left( {1 + {{\left( {\ln 2} \right)}^4} + {{\rm{e}}^2}} \right) - 0$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\ln \left( {1 + {{\left( {\ln 2} \right)}^4} + {{\rm{e}}^2}} \right)$.
