Question
$\int_{}^{} {\frac{{{{\sec }^2}x\;dx}}{{\sqrt {{{\tan }^2}x + 4} }} = } $
तब $\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{\sqrt {{{\tan }^2}x + 4} }}} = \int_{}^{} {\frac{1}{{\sqrt {{t^2} + {2^2}} }}} \,dt$
$ = \log [\tan x + \sqrt {{{\tan }^2}x + 4} ] + c.$
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