The integral I = ((1+ x)^3)/x dx equals: Option 1: 1/x^4 + C, Where C is an arbitrary constant Option 2: ((1+ x)^4)/4 + C, Where C is an arbitrary constant Option 3: ( x)/x^3 + C, Where C is an arbitrary constant Option 4: ((1+ x)^3)/3 + C, Where C is an arbitrary constant

Answer: ((1+ x)^4)/4 + C, Where C is an arbitrary constant. Solution: Substitute u=1+ x, so du=dx/x. The integral becomes u^3du=u^4/4+C=((1+ x)^4)/4+C.

Calculus

Integrals

Easy

applied

The integral $I = \int \frac{(1+\log x)^3}{x}\, dx$ equals:

$\frac{1}{x^4} + C$ , Where $C$ is an arbitrary constant

$\frac{(1+\log x)^4}{4} + C$ , Where $C$ is an arbitrary constant

$\frac{\log x}{x^3} + C$ , Where $C$ is an arbitrary constant

$\frac{(1+\log x)^3}{3} + C$ , Where $C$ is an arbitrary constant

✅ Correct Option: 2

Substitute $u=1+\log x$ , so $du=\frac{dx}{x}$ . The integral becomes $\int u^3du=\frac{u^4}{4}+C=\frac{(1+\log x)^4}{4}+C$ .

2025: 15 May Shift 1

Match List-I with List-II

List-I	List-II
(A) The value of $\int_{0}^{4} \vert x\vert \, dx$ is	(I) 3
(B) The value of $\int_{-2}^{2} \vert x\vert \, dx$ is	(II) -1
(C) The value of $\int_{0}^{3} [x] \, dx$ is	(III) 8
(D) The value of $\int_{-1}^{1} [x] \, dx$ is	(IV) 4

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

For $a, b \in R$ and $a < b$ , $\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx =$

2022: 7 Aug Shift 2

$\int \frac{1}{\cos^2 x (1 + \tan x)^3} \, dx =$

2026: 14 May Shift 2

If C is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\displaystyle\int a^x \, dx$	I.	$\sin^{-1}\dfrac{x}{a} + C$
B.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}}$	II.	$\log_e \vert x + \sqrt{x^2 - a^2} \vert + C$
C.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 + x^2}}$	III.	$\dfrac{a^x}{\log_e a} + C$
D.	$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}}$	IV.	$\log_e \vert x + \sqrt{a^2 + x^2} \vert + C$

Choose the correct answer from the options given below:

2022: 30 Aug Shift 1

$\int \frac{xe^x}{(x+1)^2} dx =$

2025: 19 May Shift 1

The value of $\int_0^1 x e^x dx$ is:

The integral I=∫(1+log⁡x)3x dxI = \int \frac{(1+\log x)^3}{x}\, dxI=∫x(1+logx)3​dx equals:

Solution

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The integral $I = \int \frac{(1+\log x)^3}{x}\, dx$ equals: