Question

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

• A. $\frac{x^3}{3}+c$
• B. $\frac{x^2}{2}+c$
• C. $\frac{x^2}{3}+c$
• D. $\frac{-x^3}{3}+c$
• E. $x+c$

Answer 1

Answer: (A)

$I=\int \frac{e^{6{\log }_{e}x}-e^{-5{\log }_{e}x}}{e^{4{\log }_{e}x}-e^{3{\log }_{e}x}}dx$
$=\int \frac{x^6-x^5}{x^4-x^3}dx=\int \frac{x^5(x-1)}{x^3(x-1)}dx$
$=\int x^{2}dx=\frac{x^3}{3}+c$