Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{10}}{9}$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y,z} \right) = \left( {\frac{1}{{{y^3} + 1}},\frac{1}{{z + 1}},1} \right)$ and the parametrization ${\bf{r}}\left( t \right) = \left( {{t^3},2,{t^2}} \right)$ for $0 \le t \le 1$.
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{1}{9},\frac{1}{{{t^2} + 1}},1} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {3{t^2},0,2t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {\frac{1}{9},\frac{1}{{{t^2} + 1}},1} \right)\cdot\left( {3{t^2},0,2t} \right)dt$
$ = \left( {\frac{1}{3}{t^2} + 2t} \right)dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 \left( {\frac{1}{3}{t^2} + 2t} \right){\rm{d}}t$
$ = \left( {\left( {\frac{1}{9}{t^3} + {t^2}} \right)|_0^1} \right) = \frac{1}{9} + 1 = \frac{{10}}{9}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{10}}{9}$.
