Question

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

Answer 1

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

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

${\displaystyle =\frac{1}{2}\int \frac{dx}{\sqrt{{\left(\frac{1}{2}\right)}^2+x^2}}}$

${\displaystyle =\frac{1}{2}\log \left|x\sqrt{\frac{1}{4}+x^2}\right|C_1}$

       .....${\displaystyle [\because \int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C]}$

${\displaystyle =\frac{1}{2}\log \left|\frac{2x+\sqrt{1+4x^2}}{2}\right|+C_1}$

${\displaystyle =\frac{1}{2}\log \left|2x+\sqrt{1+4x^2}\right|-\frac{1}{2}\log 2+C_1}$

${\displaystyle =\frac{1}{2}\log |2x+\sqrt{1}+4x^2|+C}$       .... [${\displaystyle C=-\frac{1}{2}\log 2+C_1}$]