# How do you integrate #int lnx/x^2# by integration by parts method?

# int \ (lnx)/x^2 \ dx = -(1+lnx)/x+C#

We seek:

We can then apply Integration By Parts:

Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}#

Then plugging into the IBP formula:

We have:

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To integrate ( \int \frac{\ln(x)}{x^2} ) using integration by parts, you can choose ( u = \ln(x) ) and ( dv = \frac{1}{x^2} , dx ).

This yields ( du = \frac{1}{x} , dx ) and ( v = -\frac{1}{x} ).

Applying the integration by parts formula, ( \int u , dv = uv - \int v , du ), we get:

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

The second term on the right-hand side can be integrated directly. So, the final result is:

( \int \frac{\ln(x)}{x^2} , dx = -\frac{\ln(x)}{x} - \frac{1}{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.

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