Question

Integrate the function $\frac{1}{\sqrt{1+4x^2}}$

Answer 1

Let $2x=t\Rightarrow 2dx=dt$
$\Rightarrow \int \frac{1}{\sqrt{1+4x^2}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{1+t^2}}$
$=\frac{1}{2}[\log |t+\sqrt{t^2+1}|]+C,[\because \int \frac{1}{\sqrt{x^2+a^2}}dt=\log |x+\sqrt{x^2+a^2}|]$
$=\frac{1}{2}\log |2x+\sqrt{4x^2+1}|+C$