# Find # I = int \ lnx/x^2 \ dx #?

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

We seek:

We can apply integration by by parts:

Then plugging into the IBP formula:

gives us

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To solve the integral ( I = \int \frac{\ln x}{x^2} , dx ), we can use integration by parts.

Let ( u = \ln x ) and ( dv = \frac{1}{x^2} , dx ).

Differentiating ( u ) gives ( du = \frac{1}{x} , dx ) and integrating ( dv ) gives ( v = -\frac{1}{x} ).

Now apply the integration by parts formula:

[ I = \int u , dv = uv - \int v , du ]

Substitute the values of ( u ), ( v ), ( du ), and ( dv ) into the formula:

[ I = -\ln x \cdot \frac{1}{x} - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x} , dx ]

Simplify and integrate the remaining integral:

[ I = -\frac{\ln x}{x} + \int \frac{1}{x^2} , dx ]

[ I = -\frac{\ln x}{x} - \frac{1}{x} + C ]

Thus, the integral is ( I = -\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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