Question

\[ \int \frac{e^{9 \log x} - e^{8 \log x}}{e^{6 \log x}} \, dx \] is equal to :

• A. \( x + C \)
• B. \( \frac{x^2}{2} + C \)
• C. \( \frac{x^4}{4} + C \)
• D. \( \frac{x^3}{3} + C \)

Hint:

To simplify integrals involving logarithms and exponents, use the property \( e^{\log x} = x \).

Answer 1

The Correct Option is A

Using the property \( e^{\log a} = a \), we simplify the expression as follows: \[ \int \frac{e^{9 \log x} - e^{8 \log x}}{e^{6 \log x}} \, dx = \int \frac{x^9 - x^8}{x^6} \, dx = \int x^3 - x^2 \, dx \] Now, integrate term by term: \[ \int x^3 \, dx = \frac{x^4}{4}, \quad \int x^2 \, dx = \frac{x^3}{3} \] Thus, the integral becomes: \[ \frac{x^4}{4} - \frac{x^3}{3} + C \] So, the correct answer is \( x + C \).