Q88 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
If $u$ and $v$ are functions of $x$, then $\int \dfrac{1}{v^3}\left(uv\dfrac{du}{dx} - u^2\dfrac{dv}{dx}\right)dx = $
- A$\log uv + c$
- B$\log\dfrac{u}{v} + c$
- C$\dfrac{v^2}{2u^2} + c$
- Correct answerD$\dfrac{u^2}{2v^2} + c$
Solution
The given integral is $I = \int \dfrac{1}{v^3}\left(uv\dfrac{du}{dx} - u^2\dfrac{dv}{dx}\right)dx$
Rearranging the terms in the integrand, we get:
$I = \int \dfrac{u}{v}\left(\dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\right)dx$
We know that by the quotient rule of differentiation, $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}$
Substituting this into the integral:
$I = \int \dfrac{u}{v} \dfrac{d}{dx}\left(\dfrac{u}{v}\right) dx$
Let $\dfrac{u}{v} = t$, then $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) dx = dt$
$I = \int t dt$
$I = \dfrac{t^2}{2} + c$
Substituting back $t = \dfrac{u}{v}$:
$I = \dfrac{u^2}{2v^2} + c$
Answer: $\dfrac{u^2}{2v^2} + c$
Rearranging the terms in the integrand, we get:
$I = \int \dfrac{u}{v}\left(\dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\right)dx$
We know that by the quotient rule of differentiation, $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}$
Substituting this into the integral:
$I = \int \dfrac{u}{v} \dfrac{d}{dx}\left(\dfrac{u}{v}\right) dx$
Let $\dfrac{u}{v} = t$, then $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) dx = dt$
$I = \int t dt$
$I = \dfrac{t^2}{2} + c$
Substituting back $t = \dfrac{u}{v}$:
$I = \dfrac{u^2}{2v^2} + c$
Answer: $\dfrac{u^2}{2v^2} + c$
Hence, the correct answer is option: 'D'.