e^(x_e a).e^x dx is: [where C is an arbitrary constant] Option 1: e^x/(1+_e(ae)) + C Option 2: a^x/(_e(ae)) + C Option 3: (ae)^x + C Option 4: a^xe^x/(1+_e a) + C

Answer: a^xe^x/(1+_e a) + C. Solution: e^(x_e a)=a^x, so the integrand is a^xe^x=(ae)^x. (ae)^xdx=((ae)^x)/(_e(ae))+C=a^xe^x/(1+_e a)+C, since _e(ae)=1+_e a.

Calculus

Integrals

Medium

common

$\int e^{x\log_e a}.e^x dx$ is:

[where C is an arbitrary constant]

$\frac{e^x}{1+\log_e(ae)} + C$

$\frac{a^x}{\log_e(ae)} + C$

$(ae)^x + C$

$\frac{a^xe^x}{1+\log_e a} + C$

✅ Correct Option: 4

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

$\int(ae)^xdx=\frac{(ae)^x}{\log_e(ae)}+C=\frac{a^xe^x}{1+\log_e a}+C$ , since $\log_e(ae)=1+\log_e a$ .

2026: 22 May Shift 1

$I=\int(\log x)^2dx$ is equal to :(Consider $\log_e x=\log x$ )

2026: 22 May Shift 1

Match List-I with List-II

Where C is arbitrary constant.

List-I	List-II
(A) $\displaystyle\int \dfrac{dx}{x^2 - a^2} =$	(I) $\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
(B) $\displaystyle\int \dfrac{dx}{a^2 - x^2} =$	(II) $\log\left\vert x + \sqrt{x^2-a^2}\right\vert + C$
(C) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} =$	(III) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a}\right\vert + C$
(D) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} =$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{a+x}{a-x}\right\vert + C$

Choose the correct answer from the options given below:

2026: 19 May Shift 2

$\int 5^x dx$ is

(where C is an arbitrary constant)

2026: 18 May Shift 2

If $f(x)=\int e^x(\tan x+\tan^2 x)\,dx$ and $f(0)=0$ , then $f\left(\dfrac{\pi}{4}\right)$ equals:

2026: 25 May Shift 1

Integral $I = \displaystyle\int \dfrac{e^{3\log x} - 3x^4}{e^{2\log x} + x^2}\, dx$ (where $x>0$ ) is equal to

2024: 16 May Shift 1

The value of the integral $\int_{\log _{e} 2}^{\log _{e} 3} \frac{e^{2 x}-1}{e^{2 x}+1} d x$ is :

∫exlog⁡ea.exdx\int e^{x\log_e a}.e^x dx∫exloge​a.exdx is: [where C is an arbitrary constant]

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

$\int e^{x\log_e a}.e^x dx$ is:

[where C is an arbitrary constant]