Let g(x) be an antiderivative for f(x). Then In ( 1 + ( g ( x ) ) 2 ) is an antiderivate for Options: A . &nbsp 2 f ( x ) g ( x ) 1 + ( f ( x ) ) 2 B . &nbsp 2 f ( x ) g ( x ) 1 + ( g ( x ) ) 2 C . &nbsp 2 f ( x ) 1 + ( f ( x ) ) 2 D . &nbsp None

Answer: Option B : B Given ∫ f ( x ) d x = g ( x ) g ′ ( x ) = f ( x ) Now d d x ( I n ( 1 + g 2 ( x ) ) = 2 g ( x ) g ′ ( x ) 1 + g 2 ( x ) 2 f ( x ) g ( x ) 1 + g 2 ( x )

Let g(x) be an antiderivative for f(x). Then In (1+(g(x))2) is an a

Discussion Forum : Indefinite Integration

Question -

Let g(x) be an antiderivative for f(x). Then In $(1 + (g (x))^{2})$ is an antiderivate for

Options:

A .

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

B .

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

C .

\frac{2 f (x)}{1 + (f (x))^{2}}

D . None

Answer: Option B
:
B
Given

\int f (x) d x = g (x) g^{'} (x) = f (x)

Now

\frac{d}{d x} (I n (1 + g^{2} (x)) = \frac{2 g (x) g^{'} (x)}{1 + g^{2} (x)} \frac{2 f (x) g (x)}{1 + g^{2} (x)}

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Discussion Forum : Indefinite Integration

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