Question

The value of $\int \frac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}dx$ is equal to

• A. $0+C$
• B. $\frac{x^3}{3}+C$
• C. $\frac{3}{x^3}+C$
• D. $\frac{1}{x}+C$

Answer 1

Answer: (B)

Let $I=\int \frac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}dx$
$=\int \frac{x^6-x^5}{x^4-x^3}dx\left[\because e^{y\log x}=x^y\right]$
$=\int \frac{x^5(x-1)}{x^3(x-1)}dx=\int x^{2}dx$
$=\frac{x^3}{3}+C$