If - g x d x - g x - then - g x f x - f - x dx is equal toA- gx fx-gx f-x-CB- gx f2x-CC- gx f-x-CD- gx fx-C

The correct option is D 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) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If ∫ g x d x ...

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

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

Suggest Corrections

Similar questions

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

\int \frac{[f (x) g^{^{'}} (x) + g (x) f^{^{'}} (x)]}{f (x) . g (x)} [l o g f (x) + l o g g (x)] d x

is equal to

Q. If

f (x)

and

g (x)

are periodic functions with periods

7

and

11

, respectively, then the period of

F (x) = f (x) g (\frac{x}{5}) - g (x) f (\frac{x}{3})

Q. If f(x) = 2x and

g (x) = \frac{x^{2}}{2} + 1,

then which of the following can be a discontinuous function

(a) f(x) + g(x)

(b) f(x) – g(x)

(c) f(x) g(x)

(d)

\frac{g (x)}{f (x)}

e^{A}

is defined as

e^{A} = I + A + \frac{A^{2}}{2!} + \frac{A^{3}}{3!} + . . . . . = \frac{1}{2} [\begin{matrix} f (x) & g (x) g (x) & f (x) \end{matrix}]

where

A = [\begin{matrix} x & x x & x \end{matrix}]

and

0 < x < 1

, then I is an identity matrix.

\int \frac{g (x)}{f (x)} d x

is equal to

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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