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$\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} dx$ is equal to

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x2+C

x22+C

x33+C

x42+C

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detailed solution

Correct option is C

∫e6log⁡x−e5log⁡xe4log⁡x−e3log⁡xdx=∫elog⁡x6−elog⁡x5elog⁡x4−elog⁡x3dx=∫x6−x5x4−x3dx=∫x5(x−1)x3(x−1)dx=∫x2dt=x33+C∵log⁡mn=n log ⁡m

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∫e6log⁡x−e5log⁡xe4log⁡x−e3log⁡xdx is equal to

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$\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} dx$ is equal to