Answer
$$\mathbf{r}=\left[(t+1)^{3 / 2}-1\right] \mathbf{i}+\left(1-e^{-t}\right) \mathbf{j}+[1+\ln (t+1)] \mathbf{k}$$
Work Step by Step
Since
\begin{align*}
\mathbf{r}&=\int\left[\left(\frac{3}{2}(t+1)^{1 / 2}\right) \mathbf{i}+e^{-t} \mathbf{j}+\left(\frac{1}{t+1}\right) \mathbf{k}\right] d t\\
&=(t+1)^{3 / 2} \mathbf{i}-e^{-t} \mathbf{j}+\ln (t+1) \mathbf{k}+\mathbf{C}
\end{align*}
Since
\begin{align*}
\mathbf{r}(0)&=(0+1)^{3 / 2} \mathbf{i}-e^{-0} \mathbf{j}+\ln (0+1) \mathbf{k}+\mathbf{C}\\
\mathbf{k}&=(0+1)^{3 / 2} \mathbf{i}-e^{-0} \mathbf{j}+\ln (0+1) \mathbf{k}+\mathbf{C}\\
\mathbf{C}&=-\mathbf{i}+\mathbf{j}+\mathbf{k}
\end{align*}
Hence $$\mathbf{r}=\left[(t+1)^{3 / 2}-1\right] \mathbf{i}+\left(1-e^{-t}\right) \mathbf{j}+[1+\ln (t+1)] \mathbf{k}$$
