What is the indefinite integral of #ln(ln x)#?
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The indefinite integral of (\ln(\ln x)) can be found using integration by parts. Let (u = \ln(\ln x)) and (dv = dx). Then, (du = \frac{1}{\ln x} \cdot \frac{1}{x} dx) and (v = x).
Applying integration by parts formula: [ \int u , dv = uv - \int v , du ]
[ \begin{split} \int \ln(\ln x) , dx &= x \ln(\ln x) - \int x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} , dx \ &= x \ln(\ln x) - \int \frac{1}{\ln x} , dx \end{split} ]
Now, let (w = \ln x), so (dw = \frac{1}{x} , dx). Then, the integral becomes:
[ \begin{split} \int \frac{1}{\ln x} , dx &= \int \frac{1}{w} , dw \ &= \ln |w| + C \ &= \ln |\ln x| + C \end{split} ]
Substituting this back into the previous equation:
[ \begin{split} \int \ln(\ln x) , dx &= x \ln(\ln x) - \ln |\ln x| + C \ &= x \ln(\ln x) - \ln(\ln x) + C \end{split} ]
Therefore, the indefinite integral of (\ln(\ln x)) is (x \ln(\ln x) - \ln(\ln x) + C), where (C) is the constant of integration.
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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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