# How do you find the Integral #(ln(3x))^2dx#?

Try Substitution and Integration by Parts to get:

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Try something, judge whether we seem to be making progress. (Yes, really. There's no cookbook to solve every problem.)

Applying Integration by Parts gives us

Note: you can learn a lot, by checking the answer by differentiating.

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To find the integral of ( (\ln(3x))^2 ) with respect to ( x ), you can use integration by parts. Let ( u = (\ln(3x))^2 ) and ( dv = dx ). Then, ( du = \frac{2\ln(3x)}{x} dx ) and ( v = x ).

Applying integration by parts formula:

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

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

[ = x(\ln(3x))^2 - 2\int \ln(3x) , dx ]

Now, we integrate ( \ln(3x) ):

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

[ = \frac{1}{3}x\ln(3x) - \frac{1}{3}x + C ]

So, the final result is:

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

[ = x(\ln(3x))^2 - \frac{2}{3}x\ln(3x) + \frac{2}{3}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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