# How do you integrate #(ln x) ^ 2 / x ^ 2#?

We have

Integrating by parts

Integrating by parts once again

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To integrate ( \left(\ln x\right)^2 / x^2 ), you can use integration by parts. Let ( u = \ln x ) and ( dv = u^2/x^2 , dx ). Then, differentiate ( u ) to get ( du = dx/x ), and integrate ( dv ) to get ( v = -1/x ).

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

[ \int \left(\ln x\right)^2 / x^2 , dx = -\frac{\left(\ln x\right)^2}{x} + 2 \int \frac{\ln x}{x^2} , dx ]

To integrate ( \int \frac{\ln x}{x^2} , dx ), let ( w = \ln x ), which gives ( dw = dx/x ). Substituting ( w ) and ( dw ), we get:

[ 2 \int \frac{\ln x}{x^2} , dx = 2 \int w , dw = w^2 + C ]

Finally, substitute ( w = \ln x ) back to get the complete result:

[ \int \left(\ln x\right)^2 / x^2 , dx = -\frac{\left(\ln x\right)^2}{x} + 2 \left(\ln x\right)^2 + 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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