The value of dx/(1+√x) is Option 1: √x(1+√x) + C: where C is an arbitrary constant Option 2: √x + _e, 1+√x, + C: where C is an arbitrary constant Option 3: √x - _e, 1+√x, + C: where C is an arbitrary constant Option 4: 2√x - 2_e, 1+√x, + C: where C is an arbitrary constant

Answer: 2√x - 2_e, 1+√x, + C: where C is an arbitrary constant. Solution: Put x=t^2, dx=2t dt. 2t/(1+t)dt = 2(1-1/(1+t))dt = 2t-2_e, 1+t, +C. Back-substituting t=√x gives 2√x-2_e, 1+√x, +C.

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The value of $\int \frac{dx}{1+\sqrt{x}}$ is

$\sqrt{x}(1+\sqrt{x}) + C$ : where C is an arbitrary constant

$\sqrt{x} + \log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

$\sqrt{x} - \log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

$2\sqrt{x} - 2\log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

✅ Correct Option: 4

Put $x=t^2$ , $dx=2t\,dt$ .

$\int \frac{2t}{1+t}dt = 2\int\left(1-\frac{1}{1+t}\right)dt = 2t-2\log_e|1+t|+C$ .

Back-substituting $t=\sqrt{x}$ gives $2\sqrt{x}-2\log_e|1+\sqrt{x}|+C$ .

2026: 31 May Shift 2

$\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =$

2025: 3 June Shift 2

Match List-I with List-II

List-I	List-II
(A) $\int \dfrac{dx}{x^2 - 16}$	(I) $\dfrac{1}{8} \log \left\vert \dfrac{4 + x}{4 - x} \right\vert + c$ , Where C is an arbitrary constant,
(B) $\int \dfrac{dx}{x^2 + 16}$	(II) $\log \left\vert x + \sqrt{x^2 - 16} \right\vert + c$ , Where C is an arbitrary constant,
(C) $\int \dfrac{dx}{16 - x^2}$	(III) $\dfrac{1}{8} \log \left\vert \dfrac{x - 4}{x + 4} \right\vert + c$ , Where C is an arbitrary constant,
(D) $\int \dfrac{dx}{\sqrt{x^2 - 16}}$	(IV) $\dfrac{1}{4} \tan^{-1} \left( \dfrac{x}{4} \right) + c$ , Where C is an arbitrary constant,

Choose the correct answer from the options given below:

2025: 3 June Shift 2

$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$ is equal to

2025: 2 June Shift 1

$\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$ is equal to

2026: 31 May Shift 2

Match the LIST-I with LIST-II (where $[x]$ denotes greatest integer function)

	LIST-I Integral		LIST-II Values
A.	$\displaystyle\int_{0}^{2} [x]\, dx$	I.	$0$
B.	$\displaystyle\int_{0}^{1.5} [x^2]\, dx$	II.	$1$
C.	$\displaystyle\int_{-\pi/4}^{\pi/4} \sin x \, dx$	III.	$\sqrt{2}$
D.	$\displaystyle\int_{-\pi/4}^{\pi/4} \cos x \, dx$	IV.	$2-\sqrt{2}$

Choose the correct answer from the options given below:

2026: 19 May Shift 1

$\int \frac{dx}{(a^2-x^2)^{3/2}}$ is equal to: (where C is an arbitrary constant)

The value of ∫dx1+x\int \frac{dx}{1+\sqrt{x}}∫1+x​dx​ is

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The value of $\int \frac{dx}{1+\sqrt{x}}$ is