CUET UG 2025 — Mathematics Calculus

Looking at the expression inside: $\frac{2x-1}{1+x-x^2}$1+x−x22x−1​

The formula for $\tan(A - B)$tan(A−B) is:

$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$tan(A−B)=1+tanAtanBtanA−tanB​

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Let $\tan A = x$tanA=x and $\tan B = 1-x$tanB=1−x

The numerator becomes:

$$\tan A - \tan B = x - (1-x)$$tanA−tanB=x−(1−x)

$$= x - 1 + x$$=x−1+x

$$= 2x - 1$$=2x−1

The denominator becomes:

$$1 + \tan A \tan B = 1 + x(1-x)$$1+tanAtanB=1+x(1−x)

$$= 1 + x - x^2$$=1+x−x2

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Since $\frac{2x-1}{1+x-x^2} = \frac{\tan A - \tan B}{1 + \tan A \tan B}$1+x−x22x−1​=1+tanAtanBtanA−tanB​:

$$\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right) = \tan^{-1}(x) - \tan^{-1}(1-x)$$tan−1(1+x−x22x−1​)=tan−1(x)−tan−1(1−x)

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The integral becomes:

$$\int_0^1 [\tan^{-1}(x) - \tan^{-1}(1-x)]dx$$∫01​[tan−1(x)−tan−1(1−x)]dx

$$= \int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(1-x)dx$$=∫01​tan−1(x)dx−∫01​tan−1(1−x)dx

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For $\displaystyle \int_0^1 \tan^{-1}(1-x)dx$∫01​tan−1(1−x)dx, let $u = 1-x$u=1−x

Then $du = -dx$du=−dx

When $x = 0$x=0: $u = 1$u=1

When $x = 1$x=1: $u = 0$u=0

$$\int_0^1 \tan^{-1}(1-x)dx = -\int_1^0 \tan^{-1}(u)du$$∫01​tan−1(1−x)dx=−∫10​tan−1(u)du

$$= \int_0^1 \tan^{-1}(u)du$$=∫01​tan−1(u)du

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$$\int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(1-x)dx = \int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(x)dx$$∫01​tan−1(x)dx−∫01​tan−1(1−x)dx=∫01​tan−1(x)dx−∫01​tan−1(x)dx

$$= 0$$=0

Therefore, the value of the integral is $0$0.