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$\int \frac{3 - 4 \tan x}{3 \tan x + 4} d x = \log (f (x)) + c t h e n f (x) = - - -$

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Correct option is A

∫3cosx−4sinx3sinx+4cosxdx Let 3sin⁡x+4cos⁡x=f(x)3cos⁡x−4sin⁡x=f′(x)∫f′(x)f(x)dx=log⁡(f(x))+c=log⁡(3sin⁡x+4cos⁡x)+c

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∫3−4tanx3tanx+4dx=logf(x)+c then f(x)=−−−

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$\int \frac{3 - 4 \tan x}{3 \tan x + 4} d x = \log (f (x)) + c t h e n f (x) = - - -$