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

Answer 1

Answer: (B)

$\frac{x^3}{3}$

Using power rule of logarithm $m\log a=\log a^m$

$\int \frac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}dx$ $=\int \frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}dx$

$=\int \frac{x^6-x^5}{x^4-x^3}dx$, $[\because a^{{\log }_{a}x}=x]$

$=\int \frac{x^2(x^4-x^3)}{x^4-x^3}dx$

$=\int x^{2}dx$ $\text{, cancel common factor from nr. and dr.}$

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