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$\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x=$
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The correct answer is:
$2 \sqrt{\log (\sec x+\tan x)}+c$
Let $I=\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x$
Put $\log (\sec x+\tan x)=t \Rightarrow \frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^{2} x\right) d x=d t$
$\begin{aligned} & \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \mathrm{dx}=\mathrm{dt} \Rightarrow \sec x \mathrm{dx}=\mathrm{dt} \\ \therefore \quad \mathrm{I} &=\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\int \mathrm{t}^{-\frac{1}{2}} \mathrm{dt}=\frac{\mathrm{t}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+\mathrm{c} \\ &=2 \sqrt{\mathrm{t}}+\mathrm{c}=2 \sqrt{\log (\sec x+\tan x)}+\mathrm{c} \end{aligned}$
Put $\log (\sec x+\tan x)=t \Rightarrow \frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^{2} x\right) d x=d t$
$\begin{aligned} & \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \mathrm{dx}=\mathrm{dt} \Rightarrow \sec x \mathrm{dx}=\mathrm{dt} \\ \therefore \quad \mathrm{I} &=\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\int \mathrm{t}^{-\frac{1}{2}} \mathrm{dt}=\frac{\mathrm{t}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+\mathrm{c} \\ &=2 \sqrt{\mathrm{t}}+\mathrm{c}=2 \sqrt{\log (\sec x+\tan x)}+\mathrm{c} \end{aligned}$
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