CUET UG 2025 — Mathematics Calculus

The integral to evaluate is $\displaystyle \int \frac{x}{(x-1)(x-2)} dx$∫(x−1)(x−2)x​dx

The fraction can be decomposed into partial fractions:

$$\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$$(x−1)(x−2)x​=x−1A​+x−2B​

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Multiplying both sides by $(x-1)(x-2)$(x−1)(x−2):

$$x = A(x-2) + B(x-1)$$x=A(x−2)+B(x−1)

For $x = 1$x=1:

$$1 = A(1-2) + B(0)$$1=A(1−2)+B(0)

$$1 = -A$$1=−A

$$A = -1$$A=−1

For $x = 2$x=2:

$$2 = A(0) + B(2-1)$$2=A(0)+B(2−1)

$$2 = B$$2=B

$$B = 2$$B=2

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The fraction becomes:

$$\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}$$(x−1)(x−2)x​=x−1−1​+x−22​

The integral can be written as:

$$\int \frac{x}{(x-1)(x-2)} dx = \int \frac{-1}{x-1} dx + \int \frac{2}{x-2} dx$$∫(x−1)(x−2)x​dx=∫x−1−1​dx+∫x−22​dx

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Integrating each term:

$$\int \frac{x}{(x-1)(x-2)} dx = -\ln|x-1| + 2\ln|x-2| + C$$∫(x−1)(x−2)x​dx=−ln∣x−1∣+2ln∣x−2∣+C

$$= -\ln|x-1| + \ln|(x-2)^2| + C$$=−ln∣x−1∣+ln∣(x−2)2∣+C

$$= \ln|(x-2)^2| - \ln|x-1| + C$$=ln∣(x−2)2∣−ln∣x−1∣+C

$$= \ln\left|\frac{(x-2)^2}{x-1}\right| + C$$=ln​x−1(x−2)2​​+C

Therefore, $\displaystyle \int \frac{x}{(x-1)(x-2)} dx = \ln\left|\frac{(x-2)^2}{x-1}\right| + C$∫(x−1)(x−2)x​dx=ln​x−1(x−2)2​​+C