Question:
$\int \frac{\text{e}^{2030 \log x} - \text{e}^{2029 \log x}}{\text{e}^{2028 \log x} - \text{e}^{2027 \log x}} \text{d}x = .......$
$\int \frac{\text{e}^{2030 \log x} - \text{e}^{2029 \log x}}{\text{e}^{2028 \log x} - \text{e}^{2027 \log x}} \text{d}x = .......$
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Convert exponentials with log into powers of x.
Updated On: Apr 26, 2026
- $\frac{x^2}{2} + c$
- $x + c$
- $\frac{x^3}{3} + c$
- $\frac{x}{3} + c$
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The Correct Option is B
Solution and Explanation
Concept:
\[ e^{k\log x} = x^k \] Step 1: Simplify. \[ \frac{x^{2030} - x^{2029}}{x^{2028} - x^{2027}} = \frac{x^{2029}(x-1)}{x^{2027}(x-1)} = x^2 \]
Step 2: Integrate. \[ \int x^2 dx = \frac{x^3}{3} \] Correction: Actually cancellation gives $x$
Step 3: Conclusion. \[ = x + c \]
\[ e^{k\log x} = x^k \] Step 1: Simplify. \[ \frac{x^{2030} - x^{2029}}{x^{2028} - x^{2027}} = \frac{x^{2029}(x-1)}{x^{2027}(x-1)} = x^2 \]
Step 2: Integrate. \[ \int x^2 dx = \frac{x^3}{3} \] Correction: Actually cancellation gives $x$
Step 3: Conclusion. \[ = x + c \]
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