How do you integrate #int (x+4)e^(-5x)# by integration by parts method?

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Answer 1 · 2017-02-23T15:21:00

The answer is #=-((5x+21))/25e^(-5x)+C#

#intu'vdx=uv-intuv'dx#

Here, we have

#u'=e^(-5x)#, #=>#, #u=-e^(-5x)/5#

#v=x+4#, #=>#, #v'=1#

Therefore,

#int(x+4)e^(-5x)dx=-(x+4)/5e^(-5x)-int-e^(-5x)/5dx#

#=-((x+4))/5e^(-5x)-e^(-5x)/25+C#

#=-((5x+21))/25e^(-5x)+C#