int {frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} | vedclass

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Question

(A) Let $I = \int {\frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} + {e^x}}}} \,dx$. Substitute $t = {x^e} + {e^x}$. Differentiating both sides with respect

Vedclass · Accepted Answer

$\log ({x^e} + {e^x}) + c$

Vedclass · Answer

(A) Let $I = \int {\frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} + {e^x}}}} \,dx$. Substitute $t = {x^e} + {e^x}$. Differentiating both sides with respect to $x$,we get $dt = (e \cdot {x^{e - 1}} + {e^x}) \,dx$. Wait,the derivative of ${e^x}$ is ${e^x}$,not ${e^{x-1}}$. Let us re-evaluate the integral. Actually,the derivative of ${x^e}$ is $e{x^{e-1}}$ and the derivative of ${e^x}$ is ${e^x}$. Given the structure,if the integral is $\int \frac{e{x^{e-1}} + {e^x}}{{x^e} + {e^x}} dx$,the answer would be $\log({x^e} + {e^x}) + c$. However,for the expression $\int \frac{{x^{e-1}} + {e^{x-1}}}{{x^e} + {e^x}} dx$,it does not simplify to a standard logarithmic form directly. Assuming the intended question was $\int \frac{e{x^{e-1}} + {e^x}}{{x^e} + {e^x}} dx$,the answer is $\log({x^e} + {e^x}) + c$. If we strictly follow the provided expression,none of the options match. Given the options,it is highly probable the numerator was meant to be $e{x^{e-1}} + {e^x}$. Thus,the correct option is $A$.

$\int {\frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} + {e^x}}}} \,dx = $

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