How do you use integration by parts to find #intxe^-x dx#?
1 Answer
Jul 28, 2014
#=-e^-x(1+x)+c# , where c is a constantExplanation
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 formulaSimilarly 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