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Question Number 172013 by Mikenice last updated on 23/Jun/22
find: ∫xe^(−ax) ax
find:∫xe−axax
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=∫xe−axdx{u′=e−axv=x⇒{u=−1ae−axv′=1Q=−xae−ax+1a∫e−axdx⇒Q=−xae−ax−1a2e−ax+C.

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