# What is #int ln4x^2dx#?

I tried this:

I used some properties of logs first to manipulate the integrand:

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The key to solving this integral is to know that

With this, we get

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To find the integral of ( \ln(4x^2) ) with respect to ( x ), you can use integration by parts. Let ( u = \ln(4x^2) ) and ( dv = dx ). Then, ( du = \frac{1}{x} dx ) and ( v = x ).

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

[ \int \ln(4x^2) , dx = x \ln(4x^2) - \int x \left( \frac{1}{x} \right) , dx ]

Simplify the integral:

[ = x \ln(4x^2) - \int dx ]

[ = x \ln(4x^2) - x + C ]

Where ( C ) is the constant of integration.

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