How do you find the integral #ln x / x#?

Answer 1

#int ln x / x = (ln x)^2/2 + C#

#int ln x / x = ?#

This is an example of a traditional use of u-substitution.

Let #u = ln x#. Then #du = 1/x dx#.

Currently, we have

#int ln x/x dx = int u du#

The power rule makes it simple for us to assess this:

#int u du = u^2 /2 + C#
Now, substituting back #u#, we find
#u^2 /2 + C = (ln x)^2/2 + C#

That brings us to our solution.

#int ln x / x = (ln x)^2/2 + C#
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Answer 2

To find the integral of ln(x)/x, you can use integration by parts. Let u = ln(x) and dv = dx/x. Then, differentiate u to get du and integrate dv to get v.

Integrate by parts using the formula ∫u dv = uv - ∫v du.

∫(ln(x)/x) dx = ∫u dv = uv - ∫v du = ln(x) * ln(x) - ∫(1/x * 1) dx = ln(x) * ln(x) - ∫dx/x = ln(x) * ln(x) - ln|x| + C

So, the integral of ln(x)/x is ln(x) * ln(x) - ln|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.

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.

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.

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.

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