$\int[f(x)g''(x)-f''(x)g(x)]dx$ is equal to:
Answer & explanation
Correct answer: option 3
$\int[f(x)g''(x)-f''(x)g(x)]dx=\int f(x)g''(x)dx-\int f''(x)g(x)dx$
$=(f(x)g'(x)-\int f'(x)g'(x)dx)-(g(x)f'(x)-\int g'(x)f'(x)dx)=f(x)g'(x)-f'(x)g(x)$
$\int[f(x)g''(x)-f''(x)g(x)]dx$ is equal to:
Correct answer: option 3
$\int[f(x)g''(x)-f''(x)g(x)]dx=\int f(x)g''(x)dx-\int f''(x)g(x)dx$
$=(f(x)g'(x)-\int f'(x)g'(x)dx)-(g(x)f'(x)-\int g'(x)f'(x)dx)=f(x)g'(x)-f'(x)g(x)$