Answer
$${\mathbf{r}(t) = (\arctan \mathfrak{t} +2-\frac{\pi}{4})\mathbf{i}+(1-\frac{1}{t}) \mathbf{j}+\ln t \mathbf{k}}$$
Work Step by Step
Given $$\mathbf{r}^{\prime}(t)=\frac{1}{1+t^{2}} \mathbf{i}+\frac{1}{t^{2}} \mathbf{j}+\frac{1}{t} \mathbf{k}, \quad \mathbf{r}(1)=2 \mathbf{i}$$
As we have $$\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t) dt$$
So, integrating on a component-by-component basis produces
\begin{array}{l}{\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t)(t)dt=\int\left(\frac{1}{1+t^{2}} \mathbf{i}+\frac{1}{t^{2}} \mathbf{j}+\frac{1}{t} \mathbf{k}\right) d t\\
= \arctan \mathfrak{t} \mathbf{i}-\frac{1}{t} \mathbf{j}+\ln t \mathbf{k}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(1)=2\mathbf{i} , \quad \text{so we get}
\\ {\mathbf{r}(1)= \arctan 1 \mathbf{i}-1 \mathbf{j}+\ln1 \mathbf{k} +\mathbf{C}=2\mathbf{i} }
\\ { \Rightarrow \mathbf{ C}=2\mathbf{i}- \frac{\pi}{4} \mathbf{i}+ \mathbf{j}= (2-\frac{\pi}{4})\mathbf{i}+ \mathbf{j}} \\
{\text {so we get }}\\
{\mathbf{r}(t)=\arctan \mathfrak{t} \mathbf{i}-\frac{1}{t} \mathbf{j}+\ln t \mathbf{k}+(2-\frac{\pi}{4})\mathbf{i}+ \mathbf{j}\\
= (\arctan \mathfrak{t} +2-\frac{\pi}{4})\mathbf{i}+(1-\frac{1}{t}) \mathbf{j}+\ln t \mathbf{k}}\end{array}
