x/((x-1)(x-2)) dx = (where c is an arbitrary constant) Option 1: (x^2-4x+4)/(x-1), + c Option 2: (x-1)/(x-2), + c Option 3: (x^2-2x+1)/(x-2), + c Option 4: (x-2)/(x+1), + c

Answer: (x^2-4x+4)/(x-1), + c. Solution: Partial fractions: x/((x-1)(x-2)) = (-1)/(x-1) + 2/(x-2). Integrating: -, x-1, + 2, x-2, + c =, ((x-2)^2)/(x-1), + c =, (x^2-4x+4)/(x-1), + c.

Calculus

Integrals

Medium

$\int \frac{x}{(x-1)(x-2)} dx =$

(where c is an arbitrary constant)

$\log\left|\frac{x^2-4x+4}{x-1}\right| + c$

$\log\left|\frac{x-1}{x-2}\right| + c$

$\log\left|\frac{x^2-2x+1}{x-2}\right| + c$

$\log\left|\frac{x-2}{x+1}\right| + c$

✅ Correct Option: 1

Partial fractions: $\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}$ . Integrating: $-\log|x-1| + 2\log|x-2| + c = \log\left|\frac{(x-2)^2}{x-1}\right| + c = \log\left|\frac{x^2-4x+4}{x-1}\right| + c$ .

2026: 30 May Shift 2

If $0 < x < \frac{\pi}{4}$ , then $\int\sqrt{1-\sin2x}\,dx$ is:

[where C is an arbitrary constant]

2026: 15 May Shift 1

Value of $\displaystyle\int \dfrac{dx}{\sin^{\frac{3}{5}}x \cdot \cos^{\frac{7}{5}}x}$ is (where c is an arbitrary constant)

2023: 30 May Shift 3

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$ equals :

2025: 21 May Shift 2

$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx$ is equal to:

2024: 16 May Shift 1

$\int \dfrac{\pi}{x^{n+1}-x} d x=$

2025: 14 May Shift 2

If $I_n = \int_{0}^{\pi/4} \tan^n x dx$ then $I_{2024} + I_{2026}$ is equal to:

∫x(x−1)(x−2)dx=\int \frac{x}{(x-1)(x-2)} dx =∫(x−1)(x−2)x​dx= (where c is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \frac{x}{(x-1)(x-2)} dx =$

(where c is an arbitrary constant)