If (√(x+1) + √(x-1))^2 dx = x^2 + x√(x^2-1) +, x+√(x^2-1), + C, then value of + - 2 is : Option 1: -1 Option 2: 0 Option 3: 1 Option 4: 6

Answer: 0. Solution: (√(x+1)+√(x-1))^2 = 2x + 2√(x^2-1). Integrating: x^2 + x√(x^2-1) -, x+√(x^2-1), + C. Matching gives =1, =1, =-1. Then ++2 = 0. Note: With strict reading +-2, the value is 4, which is not in the options. We have interpreted the intended expression as ++2, matching option 2.

Calculus

Integrals

Hard

If $\int(\sqrt{x+1} + \sqrt{x-1})^2 dx = \alpha x^2 + \beta x\sqrt{x^2-1} + \gamma \log|x+\sqrt{x^2-1}| + C$ , then value of $\alpha + \beta - 2\gamma$ is :

$-1$

$0$

$1$

$6$

✅ Correct Option: 2

$(\sqrt{x+1}+\sqrt{x-1})^2 = 2x + 2\sqrt{x^2-1}$ . Integrating: $x^2 + x\sqrt{x^2-1} - \log|x+\sqrt{x^2-1}| + C$ . Matching gives $\alpha=1$ , $\beta=1$ , $\gamma=-1$ . Then $\alpha+\beta+2\gamma = 0$ .

Note: With strict reading $\alpha+\beta-2\gamma$ , the value is $4$ , which is not in the options. We have interpreted the intended expression as $\alpha+\beta+2\gamma$ , matching option 2.

2026: 19 May Shift 2

$\int \frac{\log x - 3}{(\log x)^4} dx$ is equal to

(where C is an arbitrary constant and $\log_e x = \log x$ )

2026: 22 May Shift 2

$\int \frac{2\cos 2x \, dx}{(\sin x+\cos x)^2}$ is equal to:(where C is an arbitrary constant)

2025: 16 May Shift 1

$\int_{3\pi/8}^{\pi/8} \frac{ \tan^{2025} x}{ \tan^{2025} x + \cot^{2025} x} dx$ is equal to

2022: 4 Aug Shift 1

Match List I with List - II

List - I	List - II
A. An even function	I. $x^2 + \cos x$
B. For an even function, $\int_{-a}^{a} f(x)dx =$	II. 0
C. If $f(2a-x) = -f(x)$ , then $\int_{0}^{2a} f(x)dx =$	III. $2\int_{0}^{a} f(x)dx$
D. An odd function	IV. $x^3 + \sin x$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

Match List-I with List-II

(Consider $\log_e x = \log x$ )

List-I	List-II
Integral	Solution: where C is an arbitrary constant
(A) $\displaystyle\int \dfrac{f'(x)}{f(x)\log f(x)}\,dx$	(I) $x\log(\log x) + c$
(B) $\displaystyle\int \left(\log(\log x) + \dfrac{1}{\log x}\right) dx$	(II) $\dfrac{x}{\log x} + C$
(C) $\displaystyle\int \dfrac{\log x}{(1+\log x)^2}\,dx$	(III) $\log\left\vert \log(f(x)) \right\vert + c$
(D) $\displaystyle\int \left(\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right) dx$	(IV) $\dfrac{x}{1+\log x} + c$

Choose the correct answer from the options given below:

2026: 21 May Shift 1

$\int \frac{dx}{\sin^2 x \cos^2 x} =$

If ∫(x+1+x−1)2dx=αx2+βxx2−1+γlog⁡∣x+x2−1∣+C\int(\sqrt{x+1} + \sqrt{x-1})^2 dx = \alpha x^2 + \beta x\sqrt{x^2-1} + \gamma \log|x+\sqrt{x^2-1}| + C∫(x+1​+x−1​)2dx=αx2+βxx2−1​+γlog∣x+x2−1​∣+C, then value of α+β−2γ\alpha + \beta - 2\gammaα+β−2γ is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\int(\sqrt{x+1} + \sqrt{x-1})^2 dx = \alpha x^2 + \beta x\sqrt{x^2-1} + \gamma \log|x+\sqrt{x^2-1}| + C$ , then value of $\alpha + \beta - 2\gamma$ is :