Answer
True
Work Step by Step
Since $\frac{\partial}{\partial y}(M(x,y)\exp (\int \frac{M_y-N_x}{N(x,y)}dx))=M_y \exp(\int \frac{M_y-N_x}{N(x,y)}dx)$
and $\frac{\partial}{\partial y}(N(x,y)\exp (\int \frac{M_y-N_x}{N(x,y)}dx))\\=N_x \exp(\int \frac{M_y-N_x}{N(x,y)}dx)+N\frac{M_y-N_x}{N(x,y)}\exp (\int \frac{M_y-N_x}{N(x,y)}dx)\\=M_y\exp (\int \frac{M_y-N_x}{N(x,y)}dx)$
then $I(x)= \exp(\int \frac{M_y-N_x}{N(x,y)}dx)$
The differential equation $M(x,y)dx+N(x,y)dy=0$ is said to be exact if $\frac{M_y-N_x}{N(x,y)}$.
Hence, the statement is true.
