int {{e^{x log a}} cdot e^x , dx} is equal to | vedclass

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(B) We are given the integral $I = \int e^{x \log a} \cdot e^x \, dx$.Using the property of logarithms,$e^{x \log a} = e^{\log(a^x)} = a^x$.Substituti

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$\frac{(ae)^x}{\log(ae)} + c$

Vedclass · Answer

(B) We are given the integral $I = \int e^{x \log a} \cdot e^x \, dx$.Using the property of logarithms,$e^{x \log a} = e^{\log(a^x)} = a^x$.Substituting this into the integral,we get $I = \int a^x \cdot e^x \, dx$.This simplifies to $I = \int (ae)^x \, dx$.Using the standard integration formula $\int k^x \, dx = \frac{k^x}{\log k} + C$,where $k = ae$,we get $I = \frac{(ae)^x}{\log(ae)} + C$.

$\int {{e^{x \log a}} \cdot e^x \, dx}$ is equal to

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