How do you integrate #z(ln z)^2 dz#?

Answer 1

Try this:

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Answer 2

To integrate ( z(\ln z)^2 , dz ), you can use integration by parts. Let ( u = (\ln z)^2 ) and ( dv = z , dz ). Then, ( du = 2(\ln z)(1/z) , dz ) and ( v = (1/2)z^2 ).

Using the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int z(\ln z)^2 , dz = (1/2)z^2 (\ln z)^2 - \int (1/2)z^2 \cdot 2(\ln z)(1/z) , dz ]

[ = (1/2)z^2 (\ln z)^2 - \int z(\ln z) , dz ]

Now, integrate ( z(\ln z) ) using substitution:

Let ( w = \ln z ), then ( dw = (1/z) , dz ).

[ \int z(\ln z) , dz = \int w , dw ]

[ = (1/2)w^2 + C ]

Substitute ( w = \ln z ) back:

[ = (1/2)(\ln z)^2 + C ]

So, the final result is:

[ \int z(\ln z)^2 , dz = (1/2)z^2 (\ln z)^2 - (1/2)(\ln z)^2 + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.