int ({e^{alog x}} + {e^{xlog a}}) dx = | vedclass

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(C) We know that $e^{a \log x} = e^{\log x^a} = x^a$ and $e^{x \log a} = e^{\log a^x} = a^x$.Therefore,the integral becomes $\int (x^a + a^x) dx$.Usin

Vedclass · Accepted Answer

$\frac{x^{a+1}}{a+1} + \frac{a^x}{\log a} + c$

Vedclass · Answer

(C) We know that $e^{a \log x} = e^{\log x^a} = x^a$ and $e^{x \log a} = e^{\log a^x} = a^x$.Therefore,the integral becomes $\int (x^a + a^x) dx$.Using the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1} + c$ and $\int a^x dx = \frac{a^x}{\log a} + c$,we get:$\int (x^a + a^x) dx = \frac{x^{a+1}}{a+1} + \frac{a^x}{\log a} + c$.

$\int ({e^{a\log x}} + {e^{x\log a}}) dx = $

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