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Question Number 157456 by tounghoungko last updated on 23/Oct/21

∫ 5^(3−2x) dx =?

∫53−2xdx=?

Answered by FelipeLz last updated on 24/Oct/21

∫5^(3−2x) dx = 5^3 ∫5^(−2x) dx −2x = u dx = −(1/2)du −(5^3 /2)∫5^u du = −(5^3 /2)((5^u /(ln 5)))+c = −(5^(3−2x) /(2ln 5))+c

∫53−2xdx=53∫5−2xdx−2x=udx=−12du−532∫5udu=−532(5uln5)+c=−53−2x2ln5+c

Answered by mahdipoor last updated on 23/Oct/21

=∫e^((3−2x)ln5) dx=∫−(1/(2ln5))e^((3−2x)ln5) (−2ln5)dx= −(1/(2ln5))∫e^((3−2x)ln5) ((3−2x)ln5)′dx= −(1/(2ln5))∫e^u du=−(e^u /(2ln5))=−(e^((3−2x)ln5) /(2ln5))=−(5^(3−2x) /(2ln5))

=∫e(3−2x)ln5dx=∫−12ln5e(3−2x)ln5(−2ln5)dx=−12ln5∫e(3−2x)ln5((3−2x)ln5)′dx=−12ln5∫eudu=−eu2ln5=−e(3−2x)ln52ln5=−53−2x2ln5

Commented by tounghoungko last updated on 24/Oct/21

thx

thx

Answered by mevaa last updated on 23/Oct/21

=∫e^((3−2x)ln 5) dx =∫e^(3ln 5−2xln 5) dx =∫e^(3ln 5) ×e^(−2xln 5) dx =e^(3ln 5) ∫e^(−2xln 5) dx =−e^(3ln 5) /2ln 5∫−2ln 5e^(−2xln 5) dx =−e^(3ln 5) /2ln 5(e^(−2ln 5) )+constante

=∫e(3−2x)ln5dx=∫e3ln5−2xln5dx=∫e3ln5×e−2xln5dx=e3ln5∫e−2xln5dx=−e3ln5/2ln5∫−2ln5e−2xln5dx=−e3ln5/2ln5(e−2ln5)+constante

Answered by physicstutes last updated on 24/Oct/21

Let 5^(3−2x) = e^y ⇒ y = (3−2x) ln 5 ⇒ dy = −2 ln 5 dx hence dx = −(dy/(2 ln5 )) ⇒ ∫5^(3−2x) dx = −(1/(2 ln 5))∫e^y dy = −(1/(2 ln 5))(5^(3−2x) )+k

Let53−2x=ey⇒y=(3−2x)ln5⇒dy=−2ln5dxhencedx=−dy2ln5⇒∫53−2xdx=−12ln5∫eydy=−12ln5(53−2x)+k

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