Expanding: (x+x2−1)2=x2+(x2−1)+2xx2−1=2x2−1+2xx2−1. Integrating term by term: ∫2x2dx=32x3, ∫−1dx=−x, ∫2xx2−1dx=32(x2−1)3/2. So α=−1, β=2/3, γ=2/3. Then 3(α+β+γ)=3(−1+4/3)=1.
If ∫ (x + √x^2 - 1)^2 dx = α · x + β x^3 + γ (x^2 - 1)^ 3 2 + C, where C is arbitrary constant, then the value of 3(α + β + γ) is
CUET UG 2022 — Mathematics Calculus
If ∫(x+x2−1)2dx=α⋅x+βx3+γ(x2−1)23+C, where C is arbitrary constant, then the value of 3(α+β+γ) is
Held on 10 Aug 2022 · Verified 13 Jul 2026.
Options
- A
44
- B
23
- C
11
- D
1
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
CUET UG Calculus in other years
More CUET UG Calculus questions
The derivative of x³ + 2x² - 5x + 1 is:
Match **List-I** with **List-II** | List-I | List-II | |---|---| | **Function** | **Increasing on the interval** | | (A) $f(x) = -x^2 - 2x + 1$ | (I) $(-\infty, -1)$ | | (B) $f(x) = x^2 + 1$ | (II) $(1, \infty)$ | | (C) $f(x) = x^2 - 2x + 3$ | (III) $(-\infty, 0)$ | | (D) $f(x) = -x^2$ | (IV) $(0, \infty)$ | Choose the correct answer from the options given below:
For what value of $\alpha$, the function $f$ defined by $f(x) = \begin{cases} \alpha(x^2 - 2x + 1), & \text{if } x \leq 0 \\ 2x + 1, & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$?
$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx$ is equal to:
The equation of the tangent to the curve $y = \frac{(x - 3)}{(x - 1)(x - 2)}$ at the point, where it cuts x-axis is:
Other Mathematics topics for CUET UG
Keep practicing Calculus
Work through every CUET UG Calculus PYQ, year by year.