How do you integrate # [ln sqrt x] / x#?
The integral equals
We can rewrite using logarithm laws.
Hopefully this helps!
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To integrate (\frac{\ln(\sqrt{x})}{x}), you can use integration by parts. Let (u = \ln(\sqrt{x})) and (dv = \frac{1}{x}dx). Then, (du = \frac{1}{\sqrt{x}}\frac{1}{2x}dx) and (v = \ln(x)). Using the integration by parts formula:
[\int u , dv = uv - \int v , du]
we have:
[\int \frac{\ln(\sqrt{x})}{x} , dx = \ln(\sqrt{x}) \cdot \ln(x) - \int \ln(x) \cdot \frac{1}{\sqrt{x}}\frac{1}{2x} , dx]
Simplify the integral and continue integration by parts if necessary.
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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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