MathematicsMedium64×since 2004Q4039Let ∫2−tanx3+tanx dx=12(αx+loge∣βsinx+γcosx∣)+C\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C∫3+tanx2−tanx dx=21(αx+loge∣βsinx+γcosx∣)+C, where CCC is the constant of integration. Then α+γβ\alpha+\frac{\gamma}{\beta}α+βγ is equal to :A3B7C1D4Check answerSkip