Let g(x) be the anti-derivative of f(x). Then | vedclass

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(C) Given that $g(x)$ is the anti-derivative of $f(x)$,we have $g^{\prime}(x) = f(x)$.We need to find the function $h(x)$ such that $\int h(x) \, dx =

Vedclass · Accepted Answer

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

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(C) Given that $g(x)$ is the anti-derivative of $f(x)$,we have $g^{\prime}(x) = f(x)$.We need to find the function $h(x)$ such that $\int h(x) \, dx = \log _e(1+(g(x))^2) + c$.By the definition of an anti-derivative,$h(x) = \frac{d}{dx} [\log _e(1+(g(x))^2) + c]$.Using the chain rule,$\frac{d}{dx} [\log _e(1+(g(x))^2)] = \frac{1}{1+(g(x))^2} \cdot \frac{d}{dx} (1+(g(x))^2)$.$= \frac{1}{1+(g(x))^2} \cdot (2 g(x) \cdot g^{\prime}(x))$.Since $g^{\prime}(x) = f(x)$,we substitute this into the expression:$h(x) = \frac{2 g(x) f(x)}{1+(g(x))^2}$.Thus,the correct option is $C$.

Let $g(x)$ be the anti-derivative of $f(x)$. Then the function for which $\log _e(1+(g(x))^2)+c$ is an anti-derivative is:

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