e^(x_e 7) e^x dx is equal to : [where C is an arbitrary constant] Option 1: 7^x/(1+_e 7) + C Option 2: (7^x e^x)/(1+_e 7) + C Option 3: 7^x e^x + C Option 4: e^x/(1+ 7) + C

Answer: (7^x e^x)/(1+_e 7) + C. Solution: e^(x_e 7)=7^x, so the integrand is 7^xe^x=(7e)^x. (7e)^x dx = ((7e)^x)/(_e(7e)) = 7^xe^x/(1+_e 7)+C.

Calculus

Integrals

Medium

common

$\displaystyle\int e^{x\log_e 7} e^x\, dx$ is equal to :

[where C is an arbitrary constant]

$\dfrac{7^x}{1+\log_e 7} + C$

$\dfrac{7^x e^x}{1+\log_e 7} + C$

$7^x e^x + C$

$\dfrac{e^x}{1+\log 7} + C$

✅ Correct Option: 2

$e^{x\log_e 7}=7^x$ , so the integrand is $7^xe^x=(7e)^x$ .

$\int (7e)^x dx = \dfrac{(7e)^x}{\log_e(7e)} = \dfrac{7^xe^x}{1+\log_e 7}+C$ .

2026: 19 May Shift 1

The value of $\int_{0}^{3} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx$ is

2026: 22 May Shift 2

$\int_0^{2\pi} \cos^5 x \, dx$ is equal to

2023: 30 May Shift 3

$\int e^x \sec x (1 + \tan x) dx$ equals :

2025: 16 May Shift 1

The value of $\int_0^1 [\log x - \log(1-x)] dx$ is

2026: 21 May Shift 1

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

2026: 22 May Shift 2

$\int_{1/3}^{1} \frac{(x-x^3)^{1/3}}{x^4} dx$ equals