$\cfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$
$\cfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$
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NCERT & Exemplar
: Let $I = \cfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx$
Put $x = t$ $\Rightarrow$ ${\sec ^2}xdx = dt$
$= \int {\cfrac{{dt}}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }} = \log \left| {t + \sqrt {{t^2} + 4} } \right| + C}$
$= \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + C$
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