If int {{e^{sec x}}left( {sec x + tan x f(x) | vedclass

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(D) Given the integral equation: $\int e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) dx = e^{\sec x} f(x) + C$. Differentiating both si

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$\sec x + 	an x + \frac{1}{2}$

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(D) Given the integral equation: $\int e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) dx = e^{\sec x} f(x) + C$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) = \frac{d}{dx} (e^{\sec x} f(x))$. Applying the product rule on the right side: $e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) = e^{\sec x} \cdot \sec x 	an x \cdot f(x) + e^{\sec x} f'(x)$. Dividing both sides by $e^{\sec x}$: $\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x = \sec x 	an x f(x) + f'(x)$. Rearranging to solve for $f'(x)$: $f'(x) = \sec x + \sec x 	an x + \sec^2 x + 	an x f(x) - \sec x 	an x f(x)$. For the equation to hold,we set the terms involving $f(x)$ to cancel out or match the structure. If we assume $f(x) = \sec x + 	an x$,then $f'(x) = \sec x 	an x + \sec^2 x$. Substituting this into the equation: $\sec x + 	an x (\sec x + 	an x) + \sec x 	an x + \sec^2 x = \sec x 	an x (\sec x + 	an x) + (\sec x 	an x + \sec^2 x)$. This simplifies to an identity. Thus,$f(x) = \sec x + 	an x + c$. Comparing with the options,$f(x) = \sec x + 	an x + \frac{1}{2}$ is a valid choice.

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is

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