In general no, these problems are hard. BUT: You might want to look into totally unimodular matrices and total dual integrality but this requires additional assumptions on the matrix or the problem respectively. If you are lucky, then your problem has these properties and you can solve it efficiently.
Totally Unimodular Matrices
A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular. Equivalently, every square submatrix has determinant 0, +1 or −1. From the definition it follows that any submatrix of a totally unimodular matrix is itself totally unimodular (TU). Furthermore it follows that any TU matrix has only 0, +1 or −1 entries (the opposite is not true).
Some notable examples are network matrices, or the coefficient matrices of maximum flow and minimum cost flow problems.
These types of matrices are very benevolent for integer problems because if $A$ is TU and $b$ is integral, then linear programs of forms like $\{\min c^Tx ∣ A x ≥ b , x ≥ 0\}$ or $\{\max c^Tx ∣ A x ≤ b\}$ have integral optima, for any $c$.
TU matrices are also balanced matrices for which you can also show that they yield integral solutions. For these the assumptions on the matrix a more relaxed compared to totally unimodularity but to get integral solutions the right-hand side or the objective must be an all one vector.
Total Dual Integrality
Also, for TDI problems you'll get an integer optimal solution, but here the definition is a bit tricky to prove.
A rational system of inequalities $Ax≤b$ is totally dual integral (TDI) if, for all integral $c$, $\{\min y^Tb ∣ y≥0,y^TA=c\}$ is attained by an integral vector $\hat{y}$ whenever the optimum exists and is finite.
You can show that if a matrix $A$ is TU, then $Ax\leq b$ is TDI for all rational $b$.
As some other folks have pointed out your problem might fall into a specific problem type, for example set packing. For those problems there could be specialized approximation algorithms which guarantee a certain optimality gap.