Integrate the function: $\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Given the integral $I = \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} dx$.
Using the property $a \log b = \log b^a$ and $e^{\log x} = x$,we simplify the integrand:
$\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} = \frac{e^{\log x^5} - e^{\log x^4}}{e^{\log x^3} - e^{\log x^2}} = \frac{x^5 - x^4}{x^3 - x^2}$
Factor out the terms in the numerator and denominator:
$= \frac{x^4(x - 1)}{x^2(x - 1)}$
For $x \neq 1$ and $x \neq 0$,we can cancel $(x - 1)$ and $x^2$:
$= x^2$
Now,integrate the simplified function:
$\int x^2 dx = \frac{x^3}{3} + C$,where $C$ is the constant of integration.

Explore More

Similar Questions

If $\int(1-\cos x) \operatorname{cosec}^2 x \, dx = f(x) + c$,then $f(x)$ is equal to

$\int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $ equals

Difficult
View Solution

If $\int \frac{x}{(a+x)^5} dx = \frac{1}{k(a+x)^4}(f(x)) + c$,then $\frac{f(-a)}{ak} = $

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx =$

Find the integral of the function $\frac{1-\cos x}{1+\cos x}$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo