MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Recognise the integrand as $\frac{u}{v}$ multiplied by the derivative of $\frac{u}{v}$, so its integral is $\frac{1}{2}\left(\frac{u}{v}\right)^2 + c$.Explanation:Recall the quotient rule for differentiating $u$ and $v$ with respect to $x$:$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$Rewrite the given integrand by factoring out $\frac{u}{v}$:$$\frac{1}{v^3}\left(uv\frac{du}{dx} - u^2\frac{dv}{dx}\right) = \frac{u}{v}\left[\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\right]$$Using the quotient rule, the bracketed term is exactly $\frac{d}{dx}\left(\frac{u}{v}\right)$.Hence, the integrand becomes:$$\frac{u}{v} \cdot \frac{d}{dx}\left(\frac{u}{v}\right)$$Let $t = \frac{u}{v}$.Then the integral simplifies to:$$\int t \frac{dt}{dx}\, dx = \int t\, dt$$Evaluating this integral gives:$$\int t\, dt = \frac{t^2}{2} + c$$Substitute $t = \frac{u}{v}$ back to obtain the final result:$$\frac{1}{2}\left(\frac{u}{v}\right)^2 + c = \frac{u^2}{2v^2} + c$$Answer:$$\frac{u^2}{2v^2} + c$$Thus, option D is the correct answer.