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Question Number 172013 by Mikenice last updated on 23/Jun/22
find:  ∫xe^(−ax) ax
$${find}: \\ $$$$\int{xe}^{−{ax}} {ax} \\ $$
Answered by puissant last updated on 23/Jun/22
Q=∫xe^(−ax) dx     { ((u′=e^(−ax) )),((v=x  )) :}⇒   { ((u=−(1/a)e^(−ax) )),((v′=1)) :}  Q = −(x/a)e^(−ax) +(1/a)∫e^(−ax) dx  ⇒ Q = −(x/a)e^(−ax) −(1/a^2 )e^(−ax) +C.
$${Q}=\int{xe}^{−{ax}} {dx}\:\: \\ $$$$\begin{cases}{{u}'={e}^{−{ax}} }\\{{v}={x}\:\:}\end{cases}\Rightarrow\:\:\begin{cases}{{u}=−\frac{\mathrm{1}}{{a}}{e}^{−{ax}} }\\{{v}'=\mathrm{1}}\end{cases} \\ $$$${Q}\:=\:−\frac{{x}}{{a}}{e}^{−{ax}} +\frac{\mathrm{1}}{{a}}\int{e}^{−{ax}} {dx} \\ $$$$\Rightarrow\:{Q}\:=\:−\frac{{x}}{{a}}{e}^{−{ax}} −\frac{\mathrm{1}}{{a}^{\mathrm{2}} }{e}^{−{ax}} +{C}. \\ $$

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