Question

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

• A. \( \frac{1}{2} \log (t^2 - 1) + C \)
• B. \( 2 \log(t(t^2 - 1)) + C \)
• C. \( \frac{1}{4} \log \left( \frac{t}{t^2 - 3} \right) + C \)
• D. \( \frac{5}{2} \log \left( t + \frac{1}{t^2} \right) + C \)

Hint:

Always simplify parametric expressions before integrating — they often lead to clean results.

Answer 1

The Correct Option is A

Given \( x = \frac{t^3}{t^2 - 1},\ y = \frac{t}{t^2 - 1} \), compute \( x - 3y = \frac{t^3 - 3t}{t^2 - 1} \).
Then, using \( dx = \frac{d}{dt} \left( \frac{t^3}{t^2 - 1} \right) dt \), substitute and simplify:
The expression reduces nicely and becomes an integral of the form:
\[ \int \frac{dx}{x - 3y} = \frac{1}{2} \log (t^2 - 1) + C \]