Question

Answer 1

$$\begin{gathered} \int \frac{dx}{\sqrt{1+4x^2}} \\ =\int \frac{dx}{\sqrt{1+{\left(2x\right)}^2}} \\ \mathrm{let}2x=t \\ \Rightarrow 2dx=dt \\ \Rightarrow dx=\frac{dt}{2} \\ \mathrm{Now},\int \frac{dx}{\sqrt{1+{\left(2x\right)}^2}} \\ =\frac{1}{2}\int \frac{dt}{\sqrt{1+t^2}} \\ =\frac{1}{2}\log \left|t+\sqrt{1+t^2}\right|+C\left[\because \int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C\right] \\ =\frac{1}{2}\log \left|2x+\sqrt{1+4x^2}\right|+C \end{gathered}$$