How do I evaluate #int(ln(3x))^2 dx#?

Answer 1
Using Integration by substitution I set: #ln(3x)=t# so: #3x=e^t# #x=e^t/3# and #dx=1/3e^tdt#

Your integral becomes:

#intt^2*1/3e^tdt=#

Which can now be solved by parts (twice).

By parts you have:

#intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx#

Where:

#F(x)=intf(x)dy# #g'(x)# is the derivative of #g(x)# We can choose: #f(x)=e^t# #g(x)=t^2#

The integral becomes:

#intt^2*1/3e^tdt=1/3[e^t*t^2-inte^t2tdt]=# by parts again: #=1/3[e^t*t^2-e^t*2t+inte^t*2dt]=# #1/3[e^t*t^2-e^t*2t+2*e^t]=# #1/3e^t(t^2-2t+2)# but #ln(3x)=t# so going back to #x#: #=1/3*3x(ln^2(3x)-2ln(3x)+2)+c# #=x(ln^2(3x)-2ln(3x)+2)+c#
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Answer 2

To evaluate (\int (\ln(3x))^2 , dx), you can use integration by parts, where (u = (\ln(3x))^2) and (dv = dx). After differentiation and integration, the integral simplifies to (\frac{1}{3}x(\ln(3x))^2 - \frac{2}{3} \int \ln(3x) , dx). You can then integrate (\int \ln(3x) , dx) using substitution.

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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.

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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