int frac{e^{2030 log x}-e^{2029 log x}}{e^{20 | vedclass

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(C) Given the integral $I = \int \frac{e^{2030 \log x}-e^{2029 \log x}}{e^{2028 \log x}-e^{2027 \log x}} \,d x$. Using the property $e^{n \log x} = e^

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$\frac{x^3}{3}+c$,where $c$ is the constant of integration

Vedclass · Answer

(C) Given the integral $I = \int \frac{e^{2030 \log x}-e^{2029 \log x}}{e^{2028 \log x}-e^{2027 \log x}} \,d x$. Using the property $e^{n \log x} = e^{\log x^n} = x^n$,we can rewrite the integrand: $I = \int \frac{x^{2030} - x^{2029}}{x^{2028} - x^{2027}} \,d x$. Factor out the highest common powers from the numerator and denominator: $I = \int \frac{x^{2029}(x - 1)}{x^{2027}(x - 1)} \,d x$. Assuming $x 
eq 1$,we can cancel the term $(x - 1)$: $I = \int \frac{x^{2029}}{x^{2027}} \,d x = \int x^{2029 - 2027} \,d x = \int x^2 \,d x$. Integrating $x^2$ with respect to $x$,we get: $I = \frac{x^3}{3} + c$,where $c$ is the constant of integration.

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

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