I presume the trough has the vertex of the triangle down! Don't just try to remember formulas- think about what work means. The work done to lift a "piece of water" out of the trough is the weight of the piece times the height it must be lifted: the distance from the top base of the triangle to the piece. We can handle all "pieces of water" that have that same distance at once by thinking of a "layer of water" at a given distance from the top base. That will be a rectanglar layer having length 8 feet, depth dx, and width w, depending on what x is. That is, its volume will be 8w(x)dx and so its weight will be [itex]\gamma (8w(x)dx)[/itex] where [itex]\gamma[/itex] is the density of water (look it up) in pounds per cubic foot. That layer has to be lifted x feet to the top of the trough so the work done is [itex]8\gamma w(x)x dx[/itex]. Integrate that from 0 to the height of the triangle.
Now, determining w(x). Draw a picture: an equilateral triangle with vertex at the bottom, base at the height and a horizontal line x units below the base. The length of that line is w(x). "Similar triangles" is a good way to find its length. Another way would be to find the equations of the two lines, both through (0, 0) and one through [itex](1, \sqrt{3})[/itex], the other through [itex](-1, \sqrt{3})[/itex]. Be careful what you call "x" and what "w". Do you see where I got the [itex]\sqrt{3}[/itex]?