P3 Integrate (lnx)^2 between e and 1 Watch

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Kiril
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#1
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#1
Anyone? Please?
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Gauss
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#2
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(Original post by Kiril)
Anyone? Please?
Use the substitution

u = lnx
du/dx = 1/x
=> "xdu = dx"
=> "exp(u)du = dx"

Integral then becomes

INT[1,0] exp(u).u^2 du
= exp(u).u^2 - 2.INT[1,0] exp(u).u du
= exp(u).u^2 - 2 [ u.exp(u) - INT[1,0] exp(u) du ]
= exp(u).u^2 - 2u.exp(u) + 2.exp(u) | [1,0]
= exp(1) - 2exp(1) + 2exp(1) - 2
= exp(1) - 2

Galois.
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Kiril
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#3
Report Thread starter 14 years ago
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(Original post by Galois)
Use the substitution

u = lnx
du/dx = 1/x
=> "xdu = dx"
=> "exp(u)du = dx"

Integral then becomes

INT[1,0] exp(u).u^2 du
= exp(u).u^2 - 2.INT[1,0] exp(u).u du
= exp(u).u^2 - 2 [ u.exp(u) - INT[1,0] exp(u) du ]
= exp(u).u^2 - 2u.exp(u) + 2.exp(u) | [1,0]
= exp(1) - 2exp(1) + 2exp(1) - 2
= exp(1) - 2

Galois.
Cool, dude, thanx
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dvs
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#4
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Alternatively you could integrate it by parts directly using:

u = (lnx)^2
du/dx = (2/x)lnx

dv = 1 dx
v = x

Integral becomes:

x(lnx)^2 - 2 ∫ lnx dx
x(lnx)^2 - 2(xlnx-x)
x(lnx)^2 - 2xlnx+2x [e,1]
(e - 2a + 2a) - (0+2) = e-2
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