The parametric form of a curve is $x=\frac{t^3}{t^2-1}, y=\frac{t}{t^2-1}$, then $\int \frac{dx}{x-3y}=$

• Correct answer
  A

  $\frac{1}{2}\log \left(t^2-1\right)+C$
• B

  $2\log \left(t\left(t^2-1\right)\right)+C$
• C

  $\frac{1}{4}\log \left(\frac{t}{t^2-3}\right)+C$
• D

  $\frac{5}{2}\log \left(t+\frac{1}{t^2}\right)+C$

Hence, the correct answer is option: 'A'.