\(\int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx=\)
Answer & explanation
Correct answer: option 3
\(\int \frac{\sec^2xdx}{\sqrt{\tan^2x+4}}\)
$=\int\frac{d(\tan x)}{\sqrt{\tan^2x+4}}$
$=\log\left|\tan x+\sqrt{\tan^2x+4}\right|+C$
\(\int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx=\)
Correct answer: option 3
\(\int \frac{\sec^2xdx}{\sqrt{\tan^2x+4}}\)
$=\int\frac{d(\tan x)}{\sqrt{\tan^2x+4}}$
$=\log\left|\tan x+\sqrt{\tan^2x+4}\right|+C$