The parametric form of a curve is x = frac{t^ | vedclass

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(A) Given $x = \frac{t^3}{t^2 - 1}$ and $y = \frac{t}{t^2 - 1}$.First,calculate $x - 3y$:$x - 3y = \frac{t^3}{t^2 - 1} - 3 \left( \frac{t}{t^2 - 1} \r

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

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(A) Given $x = \frac{t^3}{t^2 - 1}$ and $y = \frac{t}{t^2 - 1}$.First,calculate $x - 3y$:$x - 3y = \frac{t^3}{t^2 - 1} - 3 \left( \frac{t}{t^2 - 1} \right) = \frac{t^3 - 3t}{t^2 - 1}$.Next,find $dx$ by differentiating $x$ with respect to $t$:$dx = \frac{d}{dt} \left( \frac{t^3}{t^2 - 1} \right) dt = \frac{(t^2 - 1)(3t^2) - t^3(2t)}{(t^2 - 1)^2} dt = \frac{3t^4 - 3t^2 - 2t^4}{(t^2 - 1)^2} dt = \frac{t^4 - 3t^2}{(t^2 - 1)^2} dt$.Now,substitute these into the integral:$\int \frac{dx}{x - 3y} = \int \frac{\frac{t^4 - 3t^2}{(t^2 - 1)^2} dt}{\frac{t^3 - 3t}{t^2 - 1}} = \int \frac{t^2(t^2 - 3)}{(t^2 - 1)^2} \cdot \frac{t^2 - 1}{t(t^2 - 3)} dt = \int \frac{t}{t^2 - 1} dt$.Let $u = t^2 - 1$,then $du = 2t dt$,so $t dt = \frac{1}{2} du$.$\int \frac{1}{2u} du = \frac{1}{2} \log|u| + C = \frac{1}{2} \log(t^2 - 1) + C$.

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

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