CUET UG 2025 — Mathematics Arithmetic

When faced with a complicated integral, checking if differentiating the answer options works faster can save time and reduce errors.

Consider the option: $\frac{x}{1 + \log_e x} + C$1+loge​xx​+C

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To find: $\frac{d}{dx}\left(\frac{x}{1 + \log_e x}\right)$dxd​(1+loge​xx​)

Let $u = x$u=x and $v = 1 + \log_e x$v=1+loge​x

Then $u' = 1$u′=1 and $v' = \frac{1}{x}$v′=x1​

$$\frac{d}{dx}\left(\frac{x}{1 + \log_e x}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}$$dxd​(1+loge​xx​)=v2v⋅u′−u⋅v′​

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

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

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

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This matches the original integrand.

Therefore, $\displaystyle \int \frac{\log_e x}{(1 + \log_e x)^2} dx = \frac{x}{1 + \log_e x} + C$∫(1+loge​x)2loge​x​dx=1+loge​xx​+C