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

Correct Answer is : 12log(t2−1)+C

\frac{1}{2} \log ( {t}^{2}-1)+C

\frac{1}{4} \log (\frac{t}{t^{2}-3})+C

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

AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

The parametric form of a curve is x=‌

t2−1

,y=‌

t2−1

, then ∫‌

x−3y

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