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How do you evaluate the integral of #int ln(sqrt x)/x#?

Answer 1

Isabella Ackerman

Use the fact that #ln(sqrtx) = ln(x^(1/2))=1/2lnx#

So we have:

#int ln(sqrt x)/x = 1/2 int lnx 1/xdx#

Because #d/dx(lnx) = 1/x#, we have

#1/2 int u du = 1/4u^2 +C#

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

Check the answer by differentiating.

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Answer 2

Charles Anctil

To evaluate the integral ∫ ln(√x)/x, we can use integration by parts. Let's denote u = ln(√x) and dv = dx/x. Then, differentiate u to get du and integrate dv to get v.

u = ln(√x)
du = (1/√x) * (1/2x) dx = (1/2x√x) dx
dv = dx/x
v = ln|x|

Now, we can apply the integration by parts formula: ∫udv = uv - ∫vdu.

∫ ln(√x)/x dx = ln|x| * ln(√x) - ∫ ln|x| * (1/2x√x) dx

This gives us:

∫ ln(√x)/x dx = ln|x| * ln(√x) - (1/2) ∫ (ln|x|)/√x dx

The integral on the right-hand side is not immediately solvable in terms of elementary functions, so it remains in this form.

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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.