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$\int [f (x) g^{''} (x) - f^{''} (x) g (x)] d x is equal to$

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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)

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detailed solution

Correct option is C

∫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)

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∫f(x)g′′(x)−f′′(x)g(x)dx is equal to

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$\int [f (x) g^{''} (x) - f^{''} (x) g (x)] d x is equal to$