Question
Evaluate the following integral:
$\int\frac{1}{\sqrt{1+4\text{x}^2}}\text{ dx}$

Answer

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

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