Question

$\int \frac{e^{5{\log }_{e}x}-e^{4{\log }_{e}x}}{e^{3{\log }_{e}x}-e^{2{\log }_{e}x}}dx=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$\int \frac{e^{5{\log }_{e}x}-e^{4{\log }_{e}x}}{e^{3{\log }_{e}x}-e^{2{\log }_{e}x}}dx$

$=\int \frac{e^{{\log }_e{x}^5}-e^{{\log }_e{x}^4}}{e^{{\log }_e{x}^3}-e^{{\log }_e{x}^2}}dx\left(b\log a=\log a^b\right)$

$=\int \frac{x^5-x^4}{x^3-x^2}dx\left[a^{{\log }_{a}f\left(x\right)}=f\left(x\right)\right]$

$=\int \frac{x^4\left(x-1\right)}{x^2\left(x-1\right)}dx$

$=\int x^{2}dx$

$=\frac{x^3}{3}+C$


$\int \frac{e^{5{\log }_{e}x}-e^{4{\log }_{e}x}}{e^{3{\log }_{e}x}-e^{2{\log }_{e}x}}dx=\underline{\frac{x^3}{3}+C}.$