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

1 Answer
Nov 3, 2016

Answer:

#=e^x+2x-e^(-x)+C#

Explanation:

Inegrating the given rational function is determined by decomposing
#" "#
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
#" "#
#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))#
#" "#
#=e^x+2x-e^(-x)+C#