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,

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

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

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

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