Question:
If $\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x$
$=e^{\sec x} f(x)+\mathrm{C}$, then a possible choice of $f(x)$ is:
Correct Option: 1
Solution:
Given,
$\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x+\tan x+\sec ^{2} x\right)\right) d x$
$=e^{\sec x} f(x)+C \quad \ldots(1)$
$\because \int e^{g(x)}\left(\left(g^{\prime}(x) f(x)\right)+f^{\prime}(x)\right) d x=e^{g(x)} \times f(x)+C$
Our comparing above equation by equation (1),
$f(x)=\int\left((\sec x \tan x)+\sec ^{2} x\right) d x$
$\therefore f(x)=\sec x+\tan x+C$
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