\[\int_{{}}^{{}}{[f(x)g''(x)-f''(x)g(x)]}\ dx\] is equal to [MP PET 2001]

\[\int_{{}}^{{}}{[f(x)g''(x)-f''(x)g(x)]}\ dx\] is equal to [

1.	\[\int_{{}}^{{}}{[f(x)g''(x)-f''(x)g(x)]}\ dx\] is equal to [MP PET 2001]
A.	\[\frac{f(x)}{g'(x)}\]
B.	\[f'(x)g(x)-f(x)g'(x)\]
C.	\[f(x)g'(x)-f'(x)g(x)\]
D.	\[f(x)g'(x)+f'(x)g(x)\]
Answer» D. \[f(x)g'(x)+f'(x)g(x)\]

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\[\int_{{}}^{{}}{[f(x)g''(x)-f''(x)g(x)]}\ dx\] is equal to [MP PET 2001]

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