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Question #ea2e9

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Feb 27, 2017

#int (2x)/e^x dx = -2e^(-x)(x+1)+C#

Explanation:

Considering that #1/e^x = e^(-x)# we can write the integral as:

#int (2x)/e^x dx = 2int xe^(-x)dx#

and integrate by parts:

#int (2x)/e^x dx = -2int xd(e^(-x)) = -2xe^(-x) +2inte^(-x)dx#

#int (2x)/e^x dx = -2xe^(-x) -2e^(-x) +C#

#int (2x)/e^x dx = -2e^(-x)(x+1)+C#

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