Answer
$$\int\left[\sec ^{2} t \ \mathbf{i}+\frac{1}{1+t^{2}} \ \ \mathbf{j}\right] d t=\tan t \ \mathbf{i}+\arctan (t) \ \mathbf{j}+\mathbf{C}$$
Work Step by Step
Given $$\int\left[\sec ^{2} t \ \mathbf{i}+\frac{1}{1+t^{2}} \ \mathbf{j}\right] d t $$
Integrating on a component-by-component basis produces
\begin{align}
I&=\int\left[\sec ^{2} t \ \mathbf{i}+\frac{1}{1+t^{2}} \mathbf{j}\right] d t=\tan t \ \mathbf{i}+\arctan (t) \ \mathbf{j}+\mathbf{C}
\end{align}
