x√(x + 2) dx is equal to : Option 1: 2/5(x+2)^(5/2) - 4/3(x+2)^(3/2) + C Option 2: 2/5(x+2)^(5/2) + 4/3(x+2)^(3/2) + C Option 3: 1/5(x+2)^(5/2) - 2/3(x+2)^(3/2) + C Option 4: 2/5(x+2)^(5/2) + 4/3(x+2)^(3/2) + C

Answer: 2/5(x+2)^(5/2) - 4/3(x+2)^(3/2) + C. Solution: Let t = √(x+2), so x = t^2 - 2 and dx = 2t dt. x√(x+2) dx = (t^2-2)* t * 2t dt = (2t^4 - 4t^2) dt = 2/5t^5 - 4/3t^3 + C. Replace t: 2/5(x+2)^(5/2) - 4/3(x+2)^(3/2) + C. Note: Options (2) and (4) appear identical in the paper; the correct expression with the minus sign matches option (1).

Calculus

Integrals

Medium

$\int x\sqrt{x + 2} \, dx$ is equal to :

$\frac{2}{5}(x+2)^{\frac{5}{2}} - \frac{4}{3}(x+2)^{\frac{3}{2}} + C$

$\frac{2}{5}(x+2)^{\frac{5}{2}} + \frac{4}{3}(x+2)^{\frac{3}{2}} + C$

$\frac{1}{5}(x+2)^{\frac{5}{2}} - \frac{2}{3}(x+2)^{\frac{3}{2}} + C$

$\frac{2}{5}(x+2)^{\frac{5}{2}} + \frac{4}{3}(x+2)^{\frac{3}{2}} + C$

✅ Correct Option: 1

Let $t = \sqrt{x+2}$ , so $x = t^2 - 2$ and $dx = 2t\,dt$ .

$\int x\sqrt{x+2}\,dx = \int (t^2-2)\cdot t \cdot 2t\,dt = \int (2t^4 - 4t^2)\,dt = \frac{2}{5}t^5 - \frac{4}{3}t^3 + C$ .

Replace $t$ : $\frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C$ .

Note: Options (2) and (4) appear identical in the paper; the correct expression with the minus sign matches option (1).

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Match List-I with List-II

[c is an arbitrary constant]

List-I	List-II
(A) $\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}\,dx$	(I) $\tan x - x + c$
(B) $\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx$	(II) $2\tan x - 3\sec x + c$
(C) $\int \sec x(\sec x + \tan x)\,dx$	(III) $\tan x + \cot x + c$
(D) $\int \frac{2-3\sin x}{\cos^2 x}\,dx$	(IV) $\tan x + \sec x + c$

Choose the correct answer from the options given below:

∫xx+2 dx\int x\sqrt{x + 2} \, dx∫xx+2​dx is equal to :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int x\sqrt{x + 2} \, dx$ is equal to :