\(\int\displaystyle\frac{e^{7\log_e x} - e^{6\log_e x}}{e^{4\log_e x} - e^{3\log_e x}}\,dx\) is equal to:
(Here, c is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x^4}{4} + c\)
Since \(e^{k\log_e x} = e^{\log_e x^k} = x^k\), the integrand is a ratio of powers of x:
\( \displaystyle \frac{e^{7\log_e x} - e^{6\log_e x}}{e^{4\log_e x} - e^{3\log_e x}} = \frac{x^7 - x^6}{x^4 - x^3} \)
\( \displaystyle = \frac{x^6(x - 1)}{x^3(x - 1)} = x^3 \)
So:
\( \displaystyle \int x^3\,dx = \frac{x^4}{4} + c \)