If - gxdx - gx- then - gx fx-f-xdx is equal to

∫ g(x){f(x)+f'(x)}dx =∫ g(x) f(x) dx + ∫ g(x)f'(x)dx =f(x) (∫ g(x)dx) ∫ (f'(x) ∫ g(x)dx)dx + ∫ g(x)f'(x)dx =f(x)g(x) ∫ g(x) f'(x)dx + ∫ g(x)f'(x)dx +C ...

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Byju's Answer

Standard XIII

Mathematics

Integration by Parts

If ∫ gxdx = g...

Question

If $\int g (x) d x = g (x)$ , then $\int g (x) {f (x) + f^{'} (x)} d x$ is equal to

$g (x) f (x) - g (x) f^{'} (x) + C$

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$g (x) f^{'} (x) + C$

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$g (x) f (x) + C$

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$g (x) f^{2} (x) + C$

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Solution

The correct option is C
$g (x) f (x) + C$

$\int g (x) {f (x) + f^{'} (x)} d x$
$= \int g (x) f (x) d x + \int g (x) f^{'} (x) d x$
$= f (x) (\int g (x) d x) - \int (f^{'} (x) \int g (x) d x) d x + \int g (x) f^{'} (x) d x$
$= f (x) g (x) - \int g (x) f^{'} (x) d x + \int g (x) f^{'} (x) d x + C$ ( $∵ \int g (x) d x = g (x)$ )
$= f (x) g (x) + C$

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If ∫g(x)dx=g(x), then ∫g(x){f(x)+f′(x)}dx is equal to

The correct option is C g(x)f(x)+C ∫g(x){f(x)+f′(x)}dx =∫g(x)f(x)dx+∫g(x)f′(x)dx =f(x)(∫g(x)dx)−∫(f′(x)∫g(x)dx)dx+∫g(x)f′(x)dx =f(x)g(x)−∫g(x)f′(x)dx+∫g(x)f′(x)dx+C ( ∵∫g(x)dx=g(x)) =f(x)g(x)+C

If $\int g (x) d x = g (x)$ , then $\int g (x) {f (x) + f^{'} (x)} d x$ is equal to