CUET UG 2025 — Mathematics Calculus

Find $\displaystyle \int \left\{ \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2} \right\} dx$∫{loge​x1​−(loge​x)21​}dx

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Consider $y = \frac{x}{\log_e x}$y=loge​xx​ and differentiate using the quotient rule.

For $y = \frac{u}{v}$y=vu​ where $u = x$u=x and $v = \log_e x$v=loge​x:

$$\frac{du}{dx} = 1$$dxdu​=1

$$\frac{dv}{dx} = \frac{1}{x}$$dxdv​=x1​

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Applying the quotient rule:

$$\frac{dy}{dx} = \frac{\log_e x \cdot (1) - x \cdot \left(\frac{1}{x}\right)}{(\log_e x)^2}$$dxdy​=(loge​x)2loge​x⋅(1)−x⋅(x1​)​

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

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Splitting the fraction:

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

$$= \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}$$=loge​x1​−(loge​x)21​

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Since $\frac{d}{dx}\left(\frac{x}{\log_e x}\right) = \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}$dxd​(loge​xx​)=loge​x1​−(loge​x)21​

Therefore:

$$\int \left\{ \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2} \right\} dx = \frac{x}{\log_e x} + c$$∫{loge​x1​−(loge​x)21​}dx=loge​xx​+c

where $c$c is an arbitrary constant.