The parametric form of a curve is x=\;{t³}{t²-1}, y=\;{t}{t²-1}, then \;dx/x-3y=

Correct Answer is : 12log⁡(t2−1)+C\;1/2(t²-1)+C21log(t2−1)+C

Correct Answer is : 12log⁡(t2−1)+C\;1/2(t²-1)+C21log(t2−1)+C

Test Index

AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

x=\;\frac{t^{3}}{t^{2}-1}, y=\;\frac{t}{t^{2}-1}

\int\;\frac{dx}{x-3y}=

Solution: 👈: Video Solution

x=\;\frac{t^{3}}{}, v=\;\frac{t}{2}

and let

I=\int\;\frac{dx}{}

x=\;\frac{t^{3}}{t^{2}-1}

On differentiating w.r.t.

t

, we get

\;\frac{dx}{dt}\;=\;\frac{(t^{2}-1)(3t^{2})-t^{3}(2t)}{(t^{2}-1)^{2}}

\Rightarrow \;\frac{dx}{dt}\;=\;\frac{3t^{4}-3t^{2}-2t^{4}}{(t^{2}-1)^{2}}

\Rightarrow dx\;=\;\frac{t^{4}-3t^{2}}{(t^{2}-1)^{2}}dt

\Rightarrow dx\;=\;\frac{t^{2}(t^{2}-3)}{(t^{2}-1)^{2}}dt

\Rightarrow \;\; I=\int\;\frac{1}{\left(\frac{t^{3}}{t^{2}-1}-\;\frac{3t}{t^{2}-1}\right)}\cdot\;\frac{t^{2}(t^{2}-3)}{(t^{2}-1)^{2}}dt

\;=\int\;\frac{t^{2}-1}{t(t^{2}-3)}\cdot\;\frac{t^{2}(t^{2}-3)}{(t^{2}-1)^{2}}dt

\;=\int\;\frac{t}{t^{2}-1}dt=\;\frac{1}{2}\int\;\frac{2t}{t^{2}-1}dt

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

Go to Question:

Prev Question

Next Question