Question #61a6c

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Answer 1 · 2017-04-12T20:09:39

Always let "u" equal the polynomial because differentiating it makes the resulting integral simpler

$intudv = uv-intvdu$

let $u = x^2+2x+1$ , then $du = 2x+2dx$
let $dv=e^(7x)dx$ , then $v = 1/7e^(7x)$

$int(x^2+2x+1)e^(7x)dx= 1/7(x^2+2x+1)e^(7x)-1/7int(2x+2)e^(7x)dx$

Repeat the integration by parts:

$int(x^2+2x+1)e^(7x)dx= 1/7(x^2+2x+1)e^(7x)-1/7int(2x+2)e^(7x)dx$

let $u = 1/7(2x+2)$ , then $du=2dx$
let $dv=e^(7x)dx$ , then $v = 1/7e^(7x)$

$int(x^2+2x+1)e^(7x)dx= 1/7(x^2+2x+1)e^(7x)-1/49(2x+2)e^(7x)+2/49inte^(7x)dx$

$int(x^2+2x+1)e^(7x)dx= 1/7(x^2+2x+1)e^(7x)-1/49(2x+2)e^(7x)+2/343e^(7x)+C$