(e^(7 x)-e^(6 x))/(e^(5 x)-e^(4 x)) dx, x>0 and x 1 is equal to (consider x = _e x) Option 1: x^5/5+C, where C is an arbitrary constant. Option 2: x^7/7+C, where C is an arbitrary constant. Option 3: x^8/8+C, where C is an arbitrary constant. Option 4: x^3/3+C, where C is an arbitrary constant.

Answer: x^3/3+C, where C is an arbitrary constant. Solution: Since e^(n x) = x^n, the integrand becomes (x^7-x^6)/(x^5-x^4) = (x^6(x-1))/(x^4(x-1)) = x^2. Integrating gives x^3/3+C.

Calculus

Integrals

Easy

applied

$\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx$ , $x>0$ and $x\ne 1$ is equal to (consider $\log x = \log_e x$ )

$\dfrac{x^5}{5}+C$ , where $C$ is an arbitrary constant.

$\dfrac{x^7}{7}+C$ , where $C$ is an arbitrary constant.

$\dfrac{x^8}{8}+C$ , where $C$ is an arbitrary constant.

$\dfrac{x^3}{3}+C$ , where $C$ is an arbitrary constant.

✅ Correct Option: 4

Since $e^{n\log x} = x^n$ , the integrand becomes $\dfrac{x^7-x^6}{x^5-x^4} = \dfrac{x^6(x-1)}{x^4(x-1)} = x^2$ .

Integrating gives $\dfrac{x^3}{3}+C$ .

2026: 14 May Shift 2

$\int \frac{x^8}{(x^3+1)^{1/3}} dx$ 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 :

2025: 3 June Shift 1

$\int \frac{\log_e x}{(1 + \log_e x)^2} dx$ is equal to

2022: 16 July Shift 2

Match List I with List II

List I	List II
A. $\int \frac{dx}{x^2 - a^2}$ is equal to	I. $\frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$
B. $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is equal to	II. $\frac{1}{2a}\log\left\lvert \frac{x-a}{x+a}\right\rvert + C$
C. $\int \sqrt{a^2 - x^2} \, dx$ is equal to	III. $\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$
D. $\int \sqrt{a^2 + x^2} \, dx$ is equal to	IV. $\log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$

Choose the correct answer from the options given below:

2025: 15 May Shift 2

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|) dx$ is

2023: 22 May Shift 3

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

∫e7log⁡x−e6log⁡xe5log⁡x−e4log⁡x dx\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx∫e5logx−e4logxe7logx−e6logx​dx, x>0x>0x>0 and x≠1x\ne 1x=1 is equal to (consider log⁡x=log⁡ex\log x = \log_e xlogx=loge​x)

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

$\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx$ , $x>0$ and $x\ne 1$ is equal to (consider $\log x = \log_e x$ )