(e^(7_e x) - e^(6_e x))/(e^(4_e x) - e^(3_e x)) dx is equal to: (Here, c is an arbitrary constant) Option 1: -x/2 + c Option 2: x^2/2 + c Option 3: x^3/3 + c Option 4: x^4/4 + c

Answer: x^4/4 + c. Solution: Solution Coming Soon

Calculus

Integrals

Medium

common

$\int \frac{e^{7\log_e x} - e^{6\log_e x}}{e^{4\log_e x} - e^{3\log_e x}} dx$ is equal to:

(Here, c is an arbitrary constant)

$-\frac{x}{2} + c$

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

$\frac{x^3}{3} + c$

$\frac{x^4}{4} + c$

✅ Correct Option: 4

Solution Coming Soon

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:

2026: 25 May Shift 1

$\displaystyle\int_{-\pi/3}^{\pi/3} 2x^5\cos^4 x\, dx$ equals

2022: 23 Aug Shift 1

$\int \frac{x^2 - 4}{(x+2)(x-1)(x-3)} dx =$

2026: 22 May Shift 2

$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)

2026: 30 May Shift 2

Match List-I with List-II

Choose the correct answer from the options given below:

2023: 23 May Shift 3

The value of integral $\int \sqrt{4x^2 + 9}\, dx$ is