Answer
$A(t)dt=\begin{bmatrix}
-5t & \arctan t & \frac{e^{3t}}{3}\\
\end{bmatrix} $
Work Step by Step
Given: $A(t)=\begin{bmatrix}
-5 & \frac{1}{t^2+1} & e^{3t}\\
\end{bmatrix}$
The antiderivative of the matrix function is given by:
$\int^b_aA(t)dt=\int^b_aa_{ij}(t)dt$
Hence here, $A(t)dt=\int\begin{bmatrix}-5 & \frac{1}{t^2+1} & e^{3t}
\end{bmatrix}dt=\begin{bmatrix}
-5t & \arctan t & \frac{e^{3t}}{3}\\
\end{bmatrix} $
