CUET UG 2025 — Mathematics Calculus

When an integral appears complicated, checking if any candidate function has the given expression as its derivative can be efficient.

Consider $\frac{x}{\log x}$logxx​ as a candidate for the antiderivative.

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Using the quotient rule where $u = x$u=x and $v = \log x$v=logx:

$$u' = 1$$u′=1

$$v' = \frac{1}{x}$$v′=x1​

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$$\frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{\log x \cdot (1) - x \cdot \left(\frac{1}{x}\right)}{(\log x)^2}$$dxd​(logxx​)=(logx)2logx⋅(1)−x⋅(x1​)​

$$= \frac{\log x - 1}{(\log x)^2}$$=(logx)2logx−1​

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

$$= \frac{1}{\log x} - \frac{1}{(\log x)^2}$$=logx1​−(logx)21​

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This matches the integrand exactly.

Therefore: $\displaystyle \int \left(\frac{1}{\log x} - \frac{1}{(\log x)^2}\right)dx = \frac{x}{\log x} + c$∫(logx1​−(logx)21​)dx=logxx​+c