# How do you find the integral of #int lnx/x^3 dx# from 1 to infinity?

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To find the integral of ( \int_{1}^{\infty} \frac{\ln(x)}{x^3} dx ), we first need to determine if the integral converges or diverges. We do this by analyzing the behavior of the integrand as ( x ) approaches infinity.

As ( x ) approaches infinity, the natural logarithm of ( x ) grows without bound, but ( x^3 ) grows faster than ( \ln(x) ). Therefore, the integrand approaches zero as ( x ) approaches infinity. This suggests that the integral may converge.

To evaluate the integral, we can use integration by parts. Let ( u = \ln(x) ) and ( dv = \frac{1}{x^3} dx ). Then, ( du = \frac{1}{x} dx ) and ( v = -\frac{1}{2x^2} ).

Applying the integration by parts formula, we have:

[ \int \frac{\ln(x)}{x^3} dx = -\frac{\ln(x)}{2x^2} + \int \frac{1}{2x^3} dx ]

Integrating ( \frac{1}{2x^3} ), we get ( -\frac{1}{4x^2} ).

So, the integral becomes:

[ -\frac{\ln(x)}{2x^2} - \frac{1}{4x^2} ]

Now, we evaluate this expression from 1 to ( \infty ).

As ( x ) approaches infinity, both terms approach zero.

When ( x = 1 ), the expression becomes:

[ -\frac{\ln(1)}{2 \cdot 1^2} - \frac{1}{4 \cdot 1^2} = 0 - \frac{1}{4} = -\frac{1}{4} ]

So, the integral ( \int_{1}^{\infty} \frac{\ln(x)}{x^3} dx ) converges and equals ( -\frac{1}{4} ).

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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.

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