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

Answer: e^x_e( x + x) + C. Solution: Take f(x) = ( x + x); then f'(x) = x. The integral is of the form e^x(f(x)+f'(x))dx = e^xf(x) + C. So the answer is e^x( x + x) + C.

Calculus

Integrals

Medium

core

$\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx$ is equal to: (where C is an arbitrary constant)

$e^x\log_e(\tan x) + C$

$e^x\log_e(\sec x) + C$

$e^x\sec x + C$

$e^x\log_e(\sec x + \tan x) + C$

✅ Correct Option: 4

Take $f(x) = \log(\sec x + \tan x)$ ; then $f'(x) = \sec x$ .

The integral is of the form $\int e^x(f(x)+f'(x))dx = e^xf(x) + C$ .

So the answer is $e^x\log(\sec x + \tan x) + C$ .

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(Where C is an arbitrary constant)

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If $f(x)=\int \sqrt{\dfrac{x+8}{x-8}}\,dx; f(8)=24\log_e 2$ , then $f(10)$ is equal to:

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

∫ex(sec⁡x+log⁡e(sec⁡x+tan⁡x))dx\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx∫ex(secx+loge​(secx+tanx))dx is equal to: (where C is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx$ is equal to: (where C is an arbitrary constant)