∫[f(x)g′′(x)−f"(x)g(x)]dx is equal to
f(x)g′(x)
f′(x)g(x)−f(x)g′(x)
f(x)g′(x)−f′(x)g(x)
f(x)g′(x)+f′(x)g(x)
∫[f(x)g"(x)−f"g(x)]dx =∫f(x)g"(x)dx−∫f"(x)g(x)dx =(f(x)g′(x)−∫f′(x)g′(x)dx)−(g(x)f′(x)−∫g′(x)f′(x)dx) =f(x)g′(x)−f′(x)g(x)