Question
Evaluate $\int \frac{1}{4 x^2-1} d x$

Answer

$\text { Let } I =\int \frac{ d x}{4 x^2-1}$
$=\frac{1}{4} \int \frac{ d x}{x^2-\frac{1}{4}}$
$=\frac{1}{4} \int \frac{ d x}{x^2-\left(\frac{1}{2}\right)^2}$
$=\frac{1}{4} \times \frac{1}{2 \times \frac{1}{2}} \log \left|\frac{x-\frac{1}{2}}{x+\frac{1}{2}}\right|+ c$
$\therefore I =\frac{1}{4} \log \left|\frac{2 x-1}{2 x+1}\right|+ c $

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