How do you integrate #ln(x) / x# from 4 to infinity?
the integral does not converge.
for the basic integration, spot the pattern:
So
So
the problem being, the integral does not converge.
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To integrate ln(x) / x from 4 to infinity, you can use the method of integration by parts. Let's denote u = ln(x) and dv = 1/x dx. Then, du = 1/x dx and v = ln(x).
Applying the integration by parts formula, ∫u dv = uv - ∫v du, we get:
∫ln(x) / x dx = ln(x) * ln(x) - ∫ln(x) * (1/x) dx.
This simplifies to:
∫ln(x) / x dx = ln(x) * ln(x) - ∫ln(x) / x dx.
Rearranging terms, we have:
2∫ln(x) / x dx = ln(x) * ln(x).
Now, solve for ∫ln(x) / x dx:
∫ln(x) / x dx = (1/2) * ln(x) * ln(x).
Now, to evaluate this from 4 to infinity:
lim (b→∞) ∫4 to b ln(x) / x dx = lim (b→∞) [(1/2) * ln(x) * ln(x)] from 4 to b.
= lim (b→∞) [(1/2) * ln(b) * ln(b)] - [(1/2) * ln(4) * ln(4)].
As b approaches infinity, ln(b) grows unbounded. Thus, the integral converges. Therefore, the integral from 4 to infinity of ln(x) / x dx equals infinity.
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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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