(x^3 - 1)/x^2 dx is equal to Option 1: x^2/2 + x + c, Where c is constant of integration Option 2: x^2/2 - 1/x + c, Where c is constant of integration Option 3: x^2/2 - x + c, Where c is constant of integration Option 4: x^2/2 + 1/x + c, Where c is constant of integration

Answer: x^2/2 + 1/x + c, Where c is constant of integration. Solution: (x^3 - 1)/x^2 dx = (x^3/x^2 - 1/x^2) dx = (x^(3-2) - 1/x^2) dx = (x - 1/x^2) dx = (x - x^(-2)) dx Using the power rule x^n dx = (x^(n+1))/(n+1) + c: x dx = x^2/2 x^(-2) dx = (x^(-1))/(-1) = -x^(-1) = -1/x (x^3 - 1)/x^2 dx = x^2/2 - (-1/x) + c = x^2/2 + 1/x + c Therefore, the answer is x^2/2 + 1/x + c where c is the constant of integration.

Calculus

Integrals

Easy

common

$\int \frac{x^3 - 1}{x^2} dx$ is equal to

$\frac{x^2}{2} + x + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} - \frac{1}{x} + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} - x + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} + \frac{1}{x} + c$ , Where $c$ is constant of integration

✅ Correct Option: 4

$\int \frac{x^3 - 1}{x^2} dx = \int \left(\frac{x^3}{x^2} - \frac{1}{x^2}\right) dx$

$= \int \left(x^{3-2} - \frac{1}{x^2}\right) dx$

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

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

Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + c$ :

$\int x dx = \frac{x^{2}}{2}$

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

$\int \frac{x^3 - 1}{x^2} dx = \frac{x^2}{2} - \left(-\frac{1}{x}\right) + c$

$= \frac{x^2}{2} + \frac{1}{x} + c$

Therefore, the answer is $\frac{x^2}{2} + \frac{1}{x} + c$ where $c$ is the constant of integration.

2025: 15 May Shift 1

If $\int e^x\left(\frac{x-1}{(x+1)^3}\right)dx = \frac{Ae^x}{(x+1)^B} + C$ , where C is constant of integration then which of the following are correct?

(A) $A = -1$

(B) $A = 1$

(D) $B = 2$

Choose the correct answer from the options given below:

2025: 14 May Shift 2

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2025: 30 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

$\int \frac{2\cos 2x \, dx}{(\sin x+\cos x)^2}$ is equal to:(where C is an arbitrary constant)

2023: 15 June Shift 2

The integral $\int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$ equals __________.

2026: 14 May Shift 2

If $\int_0^1 \frac{1-x}{1+x} dx = A\log 2 + B$ , then the value of A and B are

∫x3−1x2dx\int \frac{x^3 - 1}{x^2} dx∫x2x3−1​dx is equal to

Solution

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$\int \frac{x^3 - 1}{x^2} dx$ is equal to