find: ∫xe^(−ax) ax


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Question Number 172013 by Mikenice last updated on 23/Jun/22

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