If (x + √(x^2 - 1))^2 dx = * x + x^3 + (x^2 - 1)^(3/2) + C, where C is arbitrary constant, then the value of 3( + + ) is Option 1: 44 Option 2: 23 Option 3: 11 Option 4: 1

Answer: 1. Solution: Expanding: (x+√(x^2-1))^2 = x^2 + (x^2-1) + 2x√(x^2-1) = 2x^2 - 1 + 2x√(x^2-1). Integrating term by term: 2x^2 dx = 2x^3/3, -1 dx = -x, 2x√(x^2-1) dx = 2/3(x^2-1)^(3/2). So = -1, = 2/3, = 2/3. Then 3(++) = 3(-1 + 4/3) = 1.

Calculus

Integrals

Medium

If $\int (x + \sqrt{x^2 - 1})^2 \, dx = \alpha \cdot x + \beta x^3 + \gamma (x^2 - 1)^{\frac{3}{2}} + C$ , where $C$ is arbitrary constant, then the value of $3(\alpha + \beta + \gamma)$ is

✅ Correct Option: 4

Expanding: $(x+\sqrt{x^2-1})^2 = x^2 + (x^2-1) + 2x\sqrt{x^2-1} = 2x^2 - 1 + 2x\sqrt{x^2-1}$ . Integrating term by term: $\int 2x^2 dx = \frac{2x^3}{3}$ , $\int -1\, dx = -x$ , $\int 2x\sqrt{x^2-1} dx = \frac{2}{3}(x^2-1)^{3/2}$ . So $\alpha = -1$ , $\beta = 2/3$ , $\gamma = 2/3$ . Then $3(\alpha+\beta+\gamma) = 3(-1 + 4/3) = 1$ .

2025: 14 May Shift 1

The integral I = $\int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$ is equal to

2025: 19 May Shift 1

The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

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: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) The value of $I=\int_0^1 \frac{1}{\sqrt{x}}dx$ is	(I) 1
(B) The value of $I=\int_0^1 (5x^2+1)dx$ is	(II) $\frac{1}{2}$
(C) The value of $I=\int_0^\infty e^{-x}dx$ is	(III) $\frac{8}{3}$
(D) The value of $I=\int_1^e \frac{\log x}{x}dx$ is	(IV) 2

Choose the correct answer from the options given below:

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
Integral	Value of the integral
(A) $\displaystyle\int_{1}^{2} e^{2\log x} \, dx$	(I) $\log 2$
(B) $\displaystyle\int_{-1}^{1} \left\vert x \right\vert \, dx$	(II) $\dfrac{1}{2}$
(C) $\displaystyle\int_{0}^{1} \log e^x \, dx$	(III) $\dfrac{7}{3}$
(D) $\displaystyle\int_{1}^{2} \dfrac{1}{x} \, dx$	(IV) $1$

Choose the correct answer from the options given below:

2022: 16 July Shift 2

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

(where c is an arbitrary constant)

If ∫(x+x2−1)2 dx=α⋅x+βx3+γ(x2−1)32+C\int (x + \sqrt{x^2 - 1})^2 \, dx = \alpha \cdot x + \beta x^3 + \gamma (x^2 - 1)^{\frac{3}{2}} + C∫(x+x2−1​)2dx=α⋅x+βx3+γ(x2−1)23​+C, where CCC is arbitrary constant, then the value of 3(α+β+γ)3(\alpha + \beta + \gamma)3(α+β+γ) is

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\int (x + \sqrt{x^2 - 1})^2 \, dx = \alpha \cdot x + \beta x^3 + \gamma (x^2 - 1)^{\frac{3}{2}} + C$ , where $C$ is arbitrary constant, then the value of $3(\alpha + \beta + \gamma)$ is