Question

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

Answer 1

`=e^x+2x-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(e^(2x)+2e^x+1)/e^xdx`
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`=int(e^(2x))/e^xdx +int(2e^x)/e^xdx+int1/e^xdx`
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`=inte^xdx+int2dx+inte^(-x)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+2x-e^(-x)+C`