CUET UG 2025 — Mathematics Calculus

To split $\frac{1}{x(x+1)}$x(x+1)1​ into simpler parts:

$$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$$x(x+1)1​=xA​+x+1B​

Multiply both sides by $x(x+1)$x(x+1):

$$1 = A(x+1) + Bx$$1=A(x+1)+Bx

Put $x = 0$x=0:

$$1 = A(0+1) + B(0)$$1=A(0+1)+B(0)

$$1 = A$$1=A

Put $x = -1$x=−1:

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

$$1 = -B$$1=−B

$$B = -1$$B=−1

Therefore:

$$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$$x(x+1)1​=x1​−x+11​

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$$\int_{1}^{2} \frac{1}{x(x+1)} dx = \int_{1}^{2} \left(\frac{1}{x} - \frac{1}{x+1}\right) dx$$∫12​x(x+1)1​dx=∫12​(x1​−x+11​)dx

$$= \left[\ln x - \ln(x+1)\right]_{1}^{2}$$=[lnx−ln(x+1)]12​

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At $x = 2$x=2: $\ln(2) - \ln(3)$ln(2)−ln(3)

At $x = 1$x=1: $\ln(1) - \ln(2)$ln(1)−ln(2)

$$= [\ln(2) - \ln(3)] - [\ln(1) - \ln(2)]$$=[ln(2)−ln(3)]−[ln(1)−ln(2)]

$$= \ln(2) - \ln(3) - \ln(1) + \ln(2)$$=ln(2)−ln(3)−ln(1)+ln(2)

$$= 2\ln(2) - \ln(3)$$=2ln(2)−ln(3)

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Using logarithm properties:

$$= \ln(2^2) - \ln(3)$$=ln(22)−ln(3)

$$= \ln(4) - \ln(3)$$=ln(4)−ln(3)

$$= \ln\left(\frac{4}{3}\right)$$=ln(34​)

Therefore, the answer is $\ln\left(\frac{4}{3}\right)$ln(34​).