CUET UG 2025 — Mathematics Calculus

Given integral: $\displaystyle \int \left(\frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}\right) dt$∫(loge​t1​−(loge​t)21​)dt

The integrand can be recognized as the derivative of $\frac{t}{\log_e t}$loge​tt​.

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Consider differentiating $\frac{t}{\log_e t}$loge​tt​ using the quotient rule.

For $y = \frac{u}{v}$y=vu​, the derivative is $\frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2}$dtdy​=v2v⋅dtdu​−u⋅dtdv​​

Here, $u = t$u=t and $v = \log_e t$v=loge​t

So $\frac{du}{dt} = 1$dtdu​=1 and $\frac{dv}{dt} = \frac{1}{t}$dtdv​=t1​

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$$\frac{d}{dt}\left(\frac{t}{\log_e t}\right) = \frac{(\log_e t)(1) - (t)\left(\frac{1}{t}\right)}{(\log_e t)^2}$$dtd​(loge​tt​)=(loge​t)2(loge​t)(1)−(t)(t1​)​

$$= \frac{\log_e t - 1}{(\log_e t)^2}$$=(loge​t)2loge​t−1​

$$= \frac{\log_e t}{(\log_e t)^2} - \frac{1}{(\log_e t)^2}$$=(loge​t)2loge​t​−(loge​t)21​

$$= \frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}$$=loge​t1​−(loge​t)21​

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This matches the given integrand exactly.

Therefore: $\displaystyle \int \left(\frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}\right) dt = \frac{t}{\log_e t} + C$∫(loge​t1​−(loge​t)21​)dt=loge​tt​+C