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What is #int ln4x^2dx#?

Answer 1

Charles Arocha

I tried this:

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

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Answer 2

Emma Aldridge

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

The key to solving this integral is to know that

#ln(4x^2) = ln 4 + ln x^2 = ln 4 + 2 ln x#.

With this, we get

#int ln(4x^2)\ dx =# #= ln 4 int dx + 2 int ln x\ dx =# #= x ln 4 + 2 int ln x\ dx#.

Next, we view the remaining integrand as #1 \cdot ln x#, and use integration by parts with #u' = 1# and #v = ln x#. With this, we get

#x ln 4 + 2 [ x ln x - int dx ] =#

#= x ln 4 + 2 x ln x - 2x + C =#

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

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Answer 3

Isaiah Allen

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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Answer from HIX Tutor

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.