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What is #int (lnx) / x^(1/2)dx#?

Answer 1

Emma Aldridge

Put this in your mathematical toolbox: #int x^n lnx dx# can be done by parts. (#u=lnx# and #dv = x^n dx#)

In this case:

#intx^(-1/2) lnx dx#

Let #u = lnx# and #dv = x^(-1/2) dx#. So that #du = 1/x dx# and #v = 2x^(1/2)#.

#uv-vdu = 2x^(1/2)lnx-2 int x^(1/2) 1/x dx#

# = 2x^(1/2)lnx-2 int x^(-1/2)dx#

# = 2x^(1/2)lnx-4x^(1/2) +C#

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Answer 2

Paisley Johnson

The integral of (ln(x)) / x^(1/2) with respect to x is equal to 4(sqrt(x))(ln(x)) - 4(sqrt(x)) + 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.