CUET UG 2025 — Mathematics Calculus

The integral to evaluate is $\displaystyle \int \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right)dx$∫(loge​x1​−(loge​x)21​)dx.

Since the integrand contains $\log_e x$loge​x terms in the denominator, consider whether $\frac{x}{\log_e x}$loge​xx​ could be the antiderivative.

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Taking the derivative of $y = \frac{x}{\log_e x}$y=loge​xx​ using the quotient rule 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​

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

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

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Splitting the fraction:

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

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

This matches the integrand exactly.

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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