Let g(x) be an antiderivative for f(x). Then | vedclass

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(B) Given that $g(x)$ is an antiderivative of $f(x)$,we have $\frac{d}{dx}(g(x)) = f(x)$.To find the function for which $\ln(1 + (g(x))^2)$ is an anti

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

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(B) Given that $g(x)$ is an antiderivative of $f(x)$,we have $\frac{d}{dx}(g(x)) = f(x)$.To find the function for which $\ln(1 + (g(x))^2)$ is an antiderivative,we differentiate it with respect to $x$ using the chain rule:$\frac{d}{dx}(\ln(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 (2g(x) \cdot g'(x))$Since $g'(x) = f(x)$,we substitute this into the expression:$= \frac{2g(x)f(x)}{1 + (g(x))^2}$Thus,$\ln(1 + (g(x))^2)$ is an antiderivative for $\frac{2f(x)g(x)}{1 + (g(x))^2}$.

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

Let $g(x)$ be an antiderivative for $f(x)$. Then $\ln(1 + (g(x))^2)$ is an antiderivative for

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