Question

$\int \frac{\text{e}^{2030 \log x} - \text{e}^{2029 \log x}}{\text{e}^{2028 \log x} - \text{e}^{2027 \log x}} \text{d}x = .......$

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

Hint:

Use e raised to n log x equals x to the power n, then cancel common factors in the fraction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(e^{n\log x}=x^n\).

Step 2: Simplify the integrand
\(\dfrac{x^{2030}-x^{2029}}{x^{2028}-x^{2027}}=\dfrac{x^{2029}(x-1)}{x^{2027}(x-1)}=x^2\).

Step 3: Integrate
\(\int x^2\,dx=\dfrac{x^3}{3}+c\).

Step 4: Compare the options
The result is \(\frac{x^3}{3}+c\), the third option.

Final Answer:
The integral equals \(\frac{x^3}{3}+c\). \[ \boxed{\text{(C) }\dfrac{x^{3}}{3}+c} \]