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The integral becomes:
\[I = \int_{0}^{1}\frac{e^x}{e^{2x} + 1}dx\]Let \(u = e^x\). Then, \(du = e^x dx\).
Change the limits of integration:
The integral becomes:
\[I = \int_{1}^{e}\frac{du}{u^{2} + 1}\]Powered by Gemini
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