-1/x(x- x)dx equals: (consider: x=_e x) Option 1: |x- x|+C, C is an arbitrary constant. Option 2: |x+ x|+C, C is an arbitrary constant. Option 3: (x-1)+ x+C, C is an arbitrary constant. Option 4: 2|(x-1)+ x|+C, C is an arbitrary constant.

The answer is '|x- x|+C, C is an arbitrary constant.'. View full solution:

CUET Mathematics - -1/x(x- x)dx equals: (consider: x=_e x) | PYQs + Solutions

CUET Mathematics 2026 30th May Shift 1

Calculus

Medium

applied

$\log|x-\log x|+C$ , C is an arbitrary constant.

$\log|x+\log x|+C$ , C is an arbitrary constant.

$(x-1)+\log x+C$ , C is an arbitrary constant.

$2\log|(x-1)+\log x|+C$ , C is an arbitrary constant.

✅ Correct Option: 1

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Match List I with List II

LIST I	LIST II
A. $\int \frac{\sin x}{1 + \cos x} \, dx$	I. $e^{\tan^{-1} x} + C$
B. $\int \frac{1}{1 - \tan x} \, dx$	II. $\log(\log x + 1) + C$
C. $\int \frac{e^{\tan^{-1} x}}{1 + x^2} \, dx$	III. $-\log\lvert 1+\cos x \rvert + C$
D. $\int \frac{1}{x + x \log x} \, dx$	IV. $\frac{x}{2} - \frac{1}{2}\log\lvert \cos x - \sin x \rvert + C$

Choose the correct answer from the options given below: