(x-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.

Answer: x- x, +C, C is an arbitrary constant. Solution: Let u=x- x, so du=(1-1x)dx=(x-1)/xdx, which is exactly the integrand's numerator over x. The integral becomes /u=, x- x, +C.

Calculus

Integrals

Medium

applied

$\int\frac{x-1}{x(x-\log x)}dx$ equals:

(consider: $\log x=\log_e x$ )

$\log|x-\log x|+C$ , 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

Let $u=x-\log x$ , so $du=\left(1-\frac1x\right)dx=\frac{x-1}{x}dx$ , which is exactly the integrand's numerator over $x$ . The integral becomes $\int\frac{du}{u}=\log|x-\log x|+C$ .

2025: 30 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

2026: 29 May Shift 2

$\int \frac{dx}{x(x^{10}+1)}$ is equal to

2026: 30 May Shift 1

If the manufacturer's marginal cost function is $100\left(\frac{1}{\sqrt{x}}+50\right)$ , then the cost involved to increase production from 25 units to 100 units is:

2022: 16 July Shift 2

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

(where c is an arbitrary constant)

2026: 18 May Shift 2

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)+f(y)=f\left(\dfrac{x+y}{1-xy}\right)$ , Where $\mathbb{R}$ is set of real numbers then $\alpha+\int_{-\alpha}^{\alpha} f(x)\,dx$ is equal to:

2026: 12 May Shift 2

The value of $\displaystyle\int_{-1}^{1}|x^4-x|\,dx$ is equal to:

∫x−1x(x−log⁡x)dx\int\frac{x-1}{x(x-\log x)}dx∫x(x−logx)x−1​dx equals: (consider: log⁡x=log⁡ex\log x=\log_e xlogx=loge​x)

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int\frac{x-1}{x(x-\log x)}dx$ equals:

(consider: $\log x=\log_e x$ )