MCQ
If $\int \frac{d x}{4 x^2-1}=A \log \left(\frac{2 x-1}{2 x+1}\right)+c,$ then $A= ........ $
  • A
    $1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{3}$
  • $\frac{1}{4}$

Answer

Correct option: D.
$\frac{1}{4}$
$\frac{1}{4}$
$\int \frac{d x}{4 x^2-1}=A \log \left(\frac{2 x-1}{2 x+1}\right)+c ...\ ($given$)$
$\therefore \int \frac{d x}{(2 x)^2-1}= A \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$\therefore \frac{1}{2(1)(2)} \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$= A \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$\therefore A =\frac{1}{4}$

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