Question

Evaluate $\int \frac{1}{\sqrt{1+4x^2}}dx$.

Answer 1

$I=\int \frac{dx}{\sqrt{1+4x^2}}$

Let $2x=t$

$2dx=dt$

Therefore,

$I=\frac{1}{2}\int \frac{dt}{\sqrt{1+t^2}}$

$I=\frac{1}{2}\log |t+\sqrt{t^2+1}|+C$ ............. $\int \frac{1}{\sqrt{a^2+x^2}}dx=\log |x+\sqrt{a^2+x^2}|+c$

$I=\frac{1}{2}\log |2x+\sqrt{4x^2+1}|+C$