$\int \frac{\sec x}{a \tan x + b} d x is equal to$

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$\frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} - \frac{x}{2} + \frac{1}{2} \tan^{- 1} (\frac{a}{b}))| + K$

$\frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} - \frac{x}{2} - \frac{1}{2} \tan^{- 1} (\frac{a}{b}))| + C$

$\frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} + \frac{x}{2} - \frac{1}{2} \tan^{- 1} (\frac{a}{b}))| + K$

$\frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} + \frac{x}{2} + \frac{1}{2} \tan^{- 1} (\frac{a}{b}))| + C$

answer is C.

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Detailed Solution

$Let J = \int \frac{\sec x}{a \tan x + b} d x$

$\begin{array}{l} = \int \frac{1}{a \sin x + b \cos x} d x \\ = \int \frac{1}{\sqrt{a^{2} + b^{2}} [\frac{a}{\sqrt{a^{2} + b^{2}}} \sin x + \frac{b}{\sqrt{a^{2} + b^{2}}} \cos x]} d x \\ Choose \sin α = \frac{a}{\sqrt{a^{2} + b^{2}}}, \cos α = \frac{b}{\sqrt{a^{2} + b^{2}}}, \tan α = \frac{a}{b} \Rightarrow α = \tan^{- 1} (\frac{a}{b}) \end{array}$

$\begin{array}{l} J = \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{1}{\sin α \sin x + \cos α \cos x} d x \\ J = \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{1}{\cos (x - α)} d x = \frac{1}{\sqrt{a^{2} + b^{2}}} \int \sec (x - α) d x \\ J = \frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} + \frac{x - α}{2})| + C \\ J = \frac{1}{\sqrt{a^{2} + b^{2}}} \log |\tan (\frac{π}{4} + \frac{x}{2} - \frac{1}{2} \tan^{- 1} (\frac{a}{b}))| + K \end{array}$