The integral $I =\int\frac{e^{5\log_ex}-e^{4\log_ex}}{e^{3\log_ex} - e^{2\log_ex}}dx$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{x^3}{3}+C$, where C is the constant of integration
I=$\int \frac{e^{5\ln x}-e^{4\ln x}}{e^{3\ln x}-e^{2\ln x}}\,dx$
$e^{k\ln x}=x^{k}$, so
I=$\int \frac{x^{5}-x^{4}}{x^{3}-x^{2}}\,dx$
I=$\int \frac{x^{4}(x-1)}{x^{2}(x-1)}\,dx$
I=$\int x^{2}\,dx$
I=$\frac{x^{3}}{3}+C$
Final answer: $\frac{x^{3}}{3}+C$