$\int[f(x)g''(x)-f''(x)g(x)]dx$ is equal to:

$\frac{f(x)}{g'(x)}$ $f'(x)g(x)-f(x)g'(x)$ $f(x)g'(x)-f'(x)g(x)$ $f(x)g'(x)+f'(x)g(x)$

$f(x)g'(x)-f'(x)g(x)$

$\int[f(x)g''(x)-f''(x)g(x)]dx=\int f(x)g''(x)dx-\int f''(x)g(x)dx$ $=(f(x)g'(x)-\int f'(x)g'(x)dx)-(g(x)f'(x)-\int g'(x)f'(x)dx)=f(x)g'(x)-f'(x)g(x)$