Question

Am157. Politehnica 2019
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Answer 1

$\displaystyle I = \int \dfrac{2-e^x}{e^x+e^{2-x}}\, dx =\\ \\ = \int\dfrac{e^x(2-e^x)}{e^{x}(e^x+e^{2-x})}\, dx=\int\dfrac{e^x(2-e^x)}{e^{2x}+e^2}\, dx \\ \\ \\e^x = t \Rightarrow e^x \, dx = dt$

$\displaystyle I = \int \dfrac{2-t}{t^2+e^2}\, dt =\int \dfrac{2}{t^2+e^2}\, dt- \int \dfrac{t}{t^2+e^2}\, dt =\\ \\ = 2\int \dfrac{1}{t^2+e^2}\, dx - \dfrac{1}{2}\int \dfrac{(t^2+e^2)'}{t^2+e^2}\, dt = \\ \\ = \dfrac{2}{e}\,\mathrm{arctg}\, \dfrac{t}{e}-\dfrac{1}{2}\ln(t^2+e^2) +C = \\ \\ = \dfrac{2}{e}\,\mathrm{arctg}\,e^{x-1} -\dfrac{1}{2}\ln (e^{2x}+e^2)+C$