Question

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

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

Hint:

For parametric curves, convert \(dx\) into \(\dfrac{dx}{dt}dt\) and express the full integrand in terms of the parameter.

Answer 1

The Correct Option is A

Step 1: Write the given parametric equations.
The given parametric equations are
\[ x=\frac{t^3}{t^2-1} \] and \[ y=\frac{t}{t^2-1} \] We need to evaluate \[ \int \frac{dx}{x-3y} \]

Step 2: Simplify \(x-3y\).
Using the given values of \(x\) and \(y\), \[ x-3y=\frac{t^3}{t^2-1}-3\left(\frac{t}{t^2-1}\right) \] \[ x-3y=\frac{t^3-3t}{t^2-1} \] \[ x-3y=\frac{t(t^2-3)}{t^2-1} \]

Step 3: Differentiate \(x\) with respect to \(t\).
Since \[ x=\frac{t^3}{t^2-1} \] Using quotient rule, \[ \frac{dx}{dt} = \frac{(t^2-1)(3t^2)-t^3(2t)}{(t^2-1)^2} \] \[ \frac{dx}{dt} = \frac{3t^2(t^2-1)-2t^4}{(t^2-1)^2} \] \[ \frac{dx}{dt} = \frac{3t^4-3t^2-2t^4}{(t^2-1)^2} \] \[ \frac{dx}{dt} = \frac{t^4-3t^2}{(t^2-1)^2} \] \[ \frac{dx}{dt} = \frac{t^2(t^2-3)}{(t^2-1)^2} \] Therefore, \[ dx=\frac{t^2(t^2-3)}{(t^2-1)^2}\,dt \]

Step 4: Substitute in the integral.
Now, \[ \int \frac{dx}{x-3y} = \int \frac{\dfrac{t^2(t^2-3)}{(t^2-1)^2}\,dt} {\dfrac{t(t^2-3)}{t^2-1}} \] Cancel the common factors: \[ = \int \frac{t^2(t^2-3)}{(t^2-1)^2} \cdot \frac{t^2-1}{t(t^2-3)} \,dt \] \[ = \int \frac{t}{t^2-1}\,dt \]

Step 5: Integrate.
Let \[ u=t^2-1 \] Then, \[ du=2t\,dt \] So, \[ t\,dt=\frac{du}{2} \] Hence, \[ \int \frac{t}{t^2-1}\,dt = \frac{1}{2}\int \frac{du}{u} \] \[ = \frac{1}{2}\log|u|+C \] Substituting back \(u=t^2-1\), \[ = \frac{1}{2}\log|t^2-1|+C \]

Step 6: Final Answer.
Therefore, \[ \int \frac{dx}{x-3y} = \boxed{\frac{1}{2}\log(t^2-1)+C} \]