# How do you integrate int (e^(2x)+2e^x+1)/(e^x)dx?

Nov 3, 2016

$= {e}^{x} + 2 x - {e}^{- x} + C$

#### Explanation:

Inegrating the given rational function is determined by decomposing
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the given fraction into partial ones
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$\int \frac{{e}^{2 x} + 2 {e}^{x} + 1}{e} ^ x \mathrm{dx}$
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$= \int \frac{{e}^{2 x}}{e} ^ x \mathrm{dx} + \int \frac{2 {e}^{x}}{e} ^ x \mathrm{dx} + \int \frac{1}{e} ^ x \mathrm{dx}$
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$= \int {e}^{x} \mathrm{dx} + \int 2 \mathrm{dx} + \int {e}^{- x} \mathrm{dx}$
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Knowing that
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color(red)(d(e^x)=e^xdx and
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color(purple)(d(e^(-x))=-e^xrArre^(-x)=-d(e^(-x))
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=intcolor(red)(d(e^x))+int2dx+intcolor(purple)(-d(e^(-x))
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$= {e}^{x} + 2 x - {e}^{- x} + C$