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)

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

Answer 2

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