The value of int frac{e^{6 log x}-e^{5 log x | vedclass

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Question

(B) Given the integral: $ I = \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} dx $Using the logarithmic property $ e^{k \log x} = x^k

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

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(B) Given the integral: $ I = \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} dx $Using the logarithmic property $ e^{k \log x} = x^k $,we can simplify the expression:$ I = \int \frac{x^6 - x^5}{x^4 - x^3} dx $Factor out the common terms in the numerator and denominator:$ I = \int \frac{x^5(x - 1)}{x^3(x - 1)} dx $Assuming $ x 
eq 1 $,we can cancel the term $ (x - 1) $:$ I = \int \frac{x^5}{x^3} dx = \int x^2 dx $Integrating $ x^2 $ with respect to $ x $ gives:$ I = \frac{x^3}{3} + c $

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

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