How do you use integration by parts to find #intxe^-x dx#?

1 Answer
Gaurav
Jul 28, 2014

#=-e^-x(1+x)+c#, where c is a constant

Explanation

Using Integration by Parts,

#int(I)(II)dx=(I)int(II)dx-int((I)'int(II)dx)dx#

where #(I)# and #(II)# are functions of #x#, and #(I)# represents which will be differentiated and #(II)# will be integrated subsequently in the above formula

Similarly following for the problem,

#=x*inte^-xdx-int((x)'inte^-xdx)dx#

#=x*e^-x/(-1)+inte^-xdx#

#=-x*e^-x+e^-x/(-1)+c#, where c is a constant

#=-x*e^-x-e^-x+c#, where c is a constant

#=-e^-x(1+x)+c#, where c is a constant

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