#### Answer

$$\pi \mathbf{i}+\frac{\pi \sqrt{3}}{4} \mathbf{k}$$

#### Work Step by Step

We evaluate the integral of the vector function as follows:
\begin{align*}
\int_{0}^{1}\left(\frac{2}{\sqrt{1-t^{2}}} \mathbf{i}+\frac{\sqrt{3}}{1+t^{2}} \mathbf{k}\right) d t &= 2 \sin ^{-1} t \bigg|_{0}^{1} \mathbf{i}+ \sqrt{3} \tan ^{-1} t \bigg|_{0}^{1} \mathbf{k}\\
&=\pi \mathbf{i}+\frac{\pi \sqrt{3}}{4} \mathbf{k}
\end{align*}