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: Substitution
Let $t = \dfrac{u}{v}$. Then $dt = \dfrac{v\,du - u\,dv}{v^2}$.

Step 2: Rewrite the integrand
Factor $uv\,du - u^2dv = u\,(v\,du - u\,dv)$. Then $\dfrac{1}{v^3}\,u\,(v\,du - u\,dv) = \dfrac{u}{v}\cdot\dfrac{v\,du - u\,dv}{v^2} = t\,dt$.

Step 3: Integrate
$\int t\,dt = \dfrac{t^2}{2} + c = \dfrac{u^2}{2v^2} + c$.

Step 4: Check
Differentiating $\frac{t^2}{2}$ gives $t\,t'$, which agrees with the expanded form. Options A and B are logarithms, and option C inverts the fraction, so they do not match.

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