$\int\frac{e^{7\log_ex} - e^{6\log_ex}}{e^{4\log_ex} - e^{3\log_ex}} dx$ is equal to:
(Here, c is an arbitrary constant)

$\frac{x^4}{4}+c$

The correct answer is Option (4) → $\frac{x^4}{4}+c$

Given integral:

$\int \frac{e^{7\log_e x} - e^{6\log_e x}}{e^{4\log_e x} - e^{3\log_e x}}\,dx$

Use the identity $e^{a\log_e x} = x^a$:

$= \int \frac{x^7 - x^6}{x^4 - x^3} \, dx$

Simplify numerator and denominator:

$= \int \frac{x^6(x - 1)}{x^3(x - 1)} \, dx$

Cancel common factor $(x - 1)$:

$= \int \frac{x^6}{x^3} \, dx = \int x^3 \, dx$

$= \frac{x^4}{4} + C$