CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int_0^1 [\log x - \log(1-x)] \, dx$∫01​[logx−log(1−x)]dx

The integral can be split into two parts:

$$\int_0^1 [\log x - \log(1-x)] \, dx = \int_0^1 \log x \, dx - \int_0^1 \log(1-x) \, dx$$∫01​[logx−log(1−x)]dx=∫01​logxdx−∫01​log(1−x)dx

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Consider the second integral: $\displaystyle \int_0^1 \log(1-x) \, dx$∫01​log(1−x)dx

Let $u = 1-x$u=1−x

Then $du = -dx$du=−dx

Change of limits:

• When $x = 0$x=0, $u = 1$u=1
• When $x = 1$x=1, $u = 0$u=0

Substituting:

$$\int_0^1 \log(1-x) \, dx = \int_1^0 \log u \cdot (-du)$$∫01​log(1−x)dx=∫10​logu⋅(−du)

$$= -\int_1^0 \log u \, du$$=−∫10​logudu

$$= \int_0^1 \log u \, du$$=∫01​logudu

$$= \int_0^1 \log x \, dx$$=∫01​logxdx

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Substituting back into the original expression:

$$\int_0^1 [\log x - \log(1-x)] \, dx = \int_0^1 \log x \, dx - \int_0^1 \log(1-x) \, dx$$∫01​[logx−log(1−x)]dx=∫01​logxdx−∫01​log(1−x)dx

$$= \int_0^1 \log x \, dx - \int_0^1 \log x \, dx$$=∫01​logxdx−∫01​logxdx

$$= 0$$=0

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Therefore, $\displaystyle \int_0^1 [\log x - \log(1-x)] \, dx = 0$∫01​[logx−log(1−x)]dx=0