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Question
Let $f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $ be a differentiable function for all x$\in$R. Then f(x) equals :
${e^{({e^{x - 1}})}}$
$2{e^{{e^x}}} - 1$
$2{e^{{e^x} - 1}} - 1$
${e^{{e^x}}} - 1$

Solution

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