CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int \frac{f'(x)}{f(x) \log_e[f(x)]} dx$∫f(x)loge​[f(x)]f′(x)​dx

Whenever $f'(x)$f′(x) appears in the numerator and $f(x)$f(x) in the denominator, substitution simplifies the integral.

Let $u = f(x)$u=f(x)

Then $du = f'(x)dx$du=f′(x)dx

The integral becomes:

$$\int \frac{1}{u \cdot \log_e(u)} du$$∫u⋅loge​(u)1​du

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Notice that $\log_e(u)$loge​(u) appears in the denominator, and the derivative of $\log_e(u)$loge​(u) is $\frac{1}{u}$u1​.

Let $t = \log_e(u)$t=loge​(u)

Then $dt = \frac{1}{u} du$dt=u1​du

The integral becomes:

$$\int \frac{1}{t} dt$$∫t1​dt

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$$\int \frac{1}{t} dt = \log_e(t) + C$$∫t1​dt=loge​(t)+C

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Substituting back $t = \log_e(u)$t=loge​(u):

$$\log_e(t) + C = \log_e(\log_e(u)) + C$$loge​(t)+C=loge​(loge​(u))+C

Substituting back $u = f(x)$u=f(x):

$$\log_e(\log_e(u)) + C = \log_e(\log_e(f(x))) + C$$loge​(loge​(u))+C=loge​(loge​(f(x)))+C

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Therefore, $\displaystyle \int \frac{f'(x)}{f(x) \log_e[f(x)]} dx = \log_e(\log_e[f(x)]) + C$∫f(x)loge​[f(x)]f′(x)​dx=loge​(loge​[f(x)])+C