Question

$\int \frac{1+{\tan }^{2}x}{\sqrt{{\tan }^{2}x+3}}$

Answer 1

$I=\int \frac{1+{\tan }^{2}x}{\sqrt{{\tan }^{2}x+3}}dx=\int \frac{{\sec }^{2}x}{\sqrt{{\tan }^{2}x+3}}dx$

Let $\tan x=t$

${\sec }^{2}xdx=dt$

$I=\int \frac{dt}{\sqrt{t^2+3}}$

$=\ln (t+\sqrt{t^2+3})+c$