`dy = (5x^2 - 4x + 3)dx`

This is already in differential form, so we can just add the integral signs:

`int dy = int(5x^2 - 4x + 3)dx`

On the left hand side, we simply have:

`int dy = int(1)dy = y` (We are integrating the constant `1` with respect to `y`.)

On the right hand side, we integrate each of the terms, one at a time:

`int(5x^2 - 4x + 3)dx`

` = 5(x^3)/3-4(x^2/2)+3x+K`


So putting it all together, we have the solution:


Please support IntMath!