Indefinite IntegrationHard
Question
If 
g(x) dx = g(x), then
g(x) {f(x) + f′ (x)} is equal to-
g(x) dx = g(x), theng(x) {f(x) + f′ (x)} is equal to- Options
A.g(x) f(x) − g(x) f′(x) + C
B.g(x) f′(x) + C
C.g(x) f(x) + C
D.g(x) f2(x) + C
g(x) dx = g(x), theng(x) {f(x) + f′ (x)} is equal to-