Question

If \(u\) and \(v\) are functions of \(x\), then \(\int \frac{1}{v^3}(uv\frac{du}{dx}-u^2\frac{dv}{dx})dx =\)

• A. \(loguv+c\)
• B. \(log\frac{u}{v}+c\)
• C. \(\frac{v^2}{2u^2}+c\)
• D. \(\frac{u^2}{2v^2}+c\)

Hint:

Spot that the integrand is the derivative of u squared over 2 v squared.

Answer 1

The Correct Option is D

Step 1: Understand the concept
We look for a function whose derivative is the integrand. The form \(\frac{u\,u'}{v^2} - \frac{u^2v'}{v^3}\) suggests a quotient with \(u^2\) and \(v^2\).

Step 2: Expand the integrand
\[ \frac{1}{v^3}\left(uv\frac{du}{dx} - u^2\frac{dv}{dx}\right) = \frac{u}{v^2}\frac{du}{dx} - \frac{u^2}{v^3}\frac{dv}{dx} \]

Step 3: Compare with a derivative
\[ \frac{d}{dx}\left(\frac{u^2}{2v^2}\right) = \frac{2u\,u'}{2v^2} - \frac{2u^2v'}{2v^3} = \frac{u\,u'}{v^2} - \frac{u^2v'}{v^3} \]

Step 4: Result
The integrand is exactly this derivative, so the integral is \(\dfrac{u^2}{2v^2} + c\), option (D). Option (B) \(\log\frac{u}{v}\) has derivative \(\frac{u'}{u} - \frac{v'}{v}\), which is different.

Final Answer:
The integral is u^2/(2 v^2) + c. This is option (D). \[ \boxed{\text{(D) }\frac{u^2}{2v^2}+c} \]