CUET UG 2025 — Mathematics Calculus

Looking at the denominator $x - \sqrt{x}$x−x​, factor out $\sqrt{x}$x​:

$$x - \sqrt{x}$$x−x​

$$= \sqrt{x} \cdot \sqrt{x} - \sqrt{x}$$=x​⋅x​−x​

$$= \sqrt{x}(\sqrt{x} - 1)$$=x​(x​−1)

The integral becomes:

$$\int \frac{dx}{\sqrt{x}(\sqrt{x} - 1)}$$∫x​(x​−1)dx​

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Let $u = \sqrt{x}$u=x​

Then $x = u^2$x=u2 and $dx = 2u \, du$dx=2udu

Substituting:

$$\int \frac{2u \, du}{u(u - 1)}$$∫u(u−1)2udu​

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Canceling $u$u from numerator and denominator:

$$\int \frac{2u \, du}{u(u - 1)}$$∫u(u−1)2udu​

$$= \int \frac{2 \, du}{u - 1}$$=∫u−12du​

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Using the standard integral formula $\displaystyle \int \frac{1}{u-1} du = \ln|u-1| + C$∫u−11​du=ln∣u−1∣+C:

$$\int \frac{2 \, du}{u - 1}$$∫u−12du​

$$= 2 \ln|u - 1| + C$$=2ln∣u−1∣+C

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Substituting back $u = \sqrt{x}$u=x​:

$$2 \ln|\sqrt{x} - 1| + C$$2ln∣x​−1∣+C

Therefore, $\displaystyle \int \frac{dx}{x - \sqrt{x}} = 2 \log_e |\sqrt{x} - 1| + C$∫x−x​dx​=2loge​∣x​−1∣+C