Question

Integrate the following functions.
$\int \frac{dx}{\sqrt{1+4x^2}}.$

Answer 1

$\int \frac{dx}{\sqrt{1+4x^2}}=\int \frac{dx}{\sqrt{4\left(\right(\frac{1}{2}{)}^2+x^2)}}=\frac{1}{2}\int \frac{dx}{\sqrt{x^2+(\frac{1}{2}{)}^2}}=\frac{1}{2}log|x+\sqrt{x^2+(\frac{1}{2}{)}^2}|+C(\therefore \int \frac{1}{\sqrt{x^2+a^2}}dx=log|x+\sqrt{x^2+a^2}|)=\frac{1}{2}log|x+\frac{\sqrt{4x^2+1}}{2}|+C=\frac{1}{2}log|2x+\sqrt{4x^2+1}|-\frac{1}{2}log2+C[\therefore log(\frac{m}{n})=logm-logn]=\frac{1}{2}log|2x+\sqrt{4x^2+1}|+C$
$(\because \frac{1}{2}$ log 2 is constant and constant +constant+C)